Concise Mathematics Class 10 ICSE Solutions for Chapter 14 - Equation of a Line

EXERCISE 14(A)

1. Find which of the following points lie on the same line, x - 2y +5 = 0:

(i) (1, 3)

Ans: For the given point (1, 3)

Consider x = 1 and y = 3, substitute the values in the given equation

$1- 2\left( 3 \right)+5=0$

$1-6+5=0$

$-5+5=0$

$0=0$

Hence, point (1, 3) is on the line because it satisfies the equation.

(ii) (0, 5)

Ans: For the given point (0, 5)

Consider x = 5 and y = 0, substitute the values in the given equation

$0-2\left( 5 \right)+5=0$

$-10+5=0$

$-5=0$

Hence, point (2, -1.5) is not on the line because it does not satisfy the equation.

(iii) (-5, 0)

Ans: For the given point (-5, 0)

Consider x = -5 and y = 0, substitute the values in the given equation

$-5-2\left( 0 \right)+5=0$

$-5+5=0$

$0=0$

Hence, point (-5, 0) is on the line because it satisfies the equation.

(iv) (5, 5)

Ans: For the given point (5, 5)

Consider x = 5 and y = 5, substitute the values in the given equation

$5-2\left( 5 \right)+5=0$

$5+10+5=0$

$0=0$

Hence, point (5, 5) is on the line because it satisfies the equation.

(v) (2, 1.5)

Ans: For the given point (2, -1.5)

Consider x = 2 and y = -1.5, substitute the values in the given equation

$2-2\left( -1.5 \right)+5=0$

$2+3+5=0$

$10=0$

Hence, point (2, -1.5) is not on the line because it does not satisfy the equation.

(vi) (-2, -5) 

Ans: For the given point (-2, -1.5)

Consider x = 2 and y = -1.5, substitute the values in the given equation

$-2-2\left( -1.5 \right)+5=0$

$-2+3+5=0$

$0=0$

Hence, point (-2, -1.5) is on the line because it satisfies the equation.

2. State true or false:

(i) The line $\dfrac{x}{2}+\dfrac{y}{3}=0$ passes through the point (2, 3)

Ans: Given equation, $\dfrac{x}{2}+\dfrac{y}{3}=0$

Now substitute x = 2 and y = 3 in the given equation,

$\dfrac{2}{2}+\dfrac{3}{3}=0$

$1+1=0$

$2=0$

RHS $\ne  $LHS

Hence the given statement is not true.

(ii) The line $\dfrac{x}{2}+\dfrac{y}{3}=0$ passes through the point (4, -6)

Ans: Given equation, $\dfrac{x}{2}+\dfrac{y}{3}=0$

Now substitute x = 4 and y = -6 in the given equation,

$\dfrac{4}{2}+\dfrac{-6}{3}=0$

$2-2=0$

$0=0$

RHS = LHS

Hence the given statement is true.

(iii) The point (8, 7) lies on the line y - 7 = 0

Ans: Given equation y - 7 = 0

Now substitute x = 8 and y = 7 in the given equation,

$7-7=0$

$0=0$

RHS = LHS

Hence the given statement is true.

(iv) The point (-3, 0) lies on the line x + 3 = 0

Ans: Given equation, x + 3 = 0

Now substitute x = -3 and y = 0 in the given equation,

$-3+3=0$

$0=0$

RHS = LHS

Hence the given statement is true.

(v) The point (2, a) lies on the line 2x - y = 3 then a = 5

Ans: Given equation, 2x - y = 3

Now substitute x = 2 and y = a in the given equation,

$2\left( 2 \right)-a=3$

$4-a=3$

$-a= -1$

RHS $\ne $ LHS

Hence the given statement is not true.

3. The line given by the equation $2x-\dfrac{y}{3}=7$ passes through the point (k, 6); calculate the value of k.

Ans: Given equation $2x-\dfrac{y}{3}=7$

Now, substitute x = k and y = 6 in the given equation,

$2k-\dfrac{6}{3}=7$

$2k-2=7$

$2k=9$

$k=\dfrac{9}{2}$

$k=4.5$

4. For what value of k will the point (3, -k) lie on the line 9x + 4y = 3?

Ans: Given equation 9x + 4y = 3

Now, substitute x = 3 and y = -k in the given equation,

$9\left( 3 \right)+4\left( -k \right)=3$

$27-4k=3$

$4k= -24$

$k=\dfrac{-24}{4}$

$k= -6$

5. The line $\dfrac{3x}{5}-\dfrac{2y}{3}+1=0$ contains the point (m, 2m - 1); calculate the value of m.

Ans: Given equation $\dfrac{3x}{5}-\dfrac{2y}{3}+1=0$

Now, substitute x = m and y = 2m - 1 in the given equation,

$\dfrac{3m}{5}-\dfrac{2\left( 2m - 1 \right)}{3}+1=0$

$\dfrac{3m}{5}-\dfrac{4m - 2}{3}+1=0$

$\dfrac{9m - 20m + 10 + 15}{15}=0$

$-11m+25=0$

$-11m = -25$

$m=\dfrac{25}{11}$

6. Does the line 3x - 5y = 6 bisect the join of (5, -2) and (-1, 2)?

Ans: Given equation of line is 3x - 5y = 6

So, the end points are (5, -2) and (-1, 2).

Now the coordinates of midpoint will be,

$\left( x,y \right)=\left( \dfrac{{{x}_{2}} + {{x}_{1}}}{2},\dfrac{{{y}_{2}} + {{y}_{1}}}{2} \right)$

$\left( x,y \right)=\left( \dfrac{5 + \left( -1 \right)}{2},\dfrac{-2 + 2}{2} \right)$

$\left( x,y \right)=\left( \dfrac{4}{2},\dfrac{0}{2} \right)$

$\left( x,y \right)=\left( 2,0 \right)$

Substitute x = 2 and y = 0 in the given equation of line.

3(2) - 5(0) = 6

6 = 6

Hence, the line 3x - 5y = 6 bisects the join of (5, -2) and (-1, 2).

7. (i) The line y = 3x - 2 bisects the join of (a, 3) and (2, -5), find the value of a. 

Ans: Given equation of line is, y = 3x - 2. Then the end points are (a, 3) and (2, -5). Now the coordinate of midpoint will be,

$\left( x,y \right)=\left( \dfrac{{{x}_{2}} + {{x}_{1}}}{2},\dfrac{{{y}_{2}} + {{y}_{1}}}{2} \right)$

$\left( x,y \right)=\left( \dfrac{a + 2}{2},\dfrac{3 + \left( -5 \right)}{2} \right)$

$\left( x,y \right)=\left( \dfrac{a + 2}{2}, -1 \right)$

Substitute $x=\dfrac{a + 2}{2}$ and $y= -1$ in the given equation of line.

$-1=3\left( \dfrac{a + 2}{2} \right)-2$

$\dfrac{1}{3}=\left( \dfrac{a + 2}{2} \right)$

$\dfrac{2}{3}=a + 2$

$a=\dfrac{2}{3}-2$

$a=\dfrac{2 - 6}{3}$

$a=\dfrac{-4}{3}$

(ii) The line x - 6y + 11 = 0 bisects the join of (8, -1) and (0, k). Find the value of k.

Ans: Given equation of line is, y = 3x - 2. Then the line joining points are (8, -1) and (0, k).

$\left( x,y \right)=\left( \dfrac{{{x}_{2}} + {{x}_{1}}}{2},\dfrac{{{y}_{2}} + {{y}_{1}}}{2} \right)$

$\left( x,y \right)=\left( \dfrac{8 + 0}{2},\dfrac{-1 + k}{2} \right)$

$\left( x,y \right)=\left( 4,\dfrac{k - 1}{2} \right)$

Substitute x = 4 and y = $\dfrac{k - 1}{2}$ in the given equation of line,

$4-6\left( \dfrac{k - 1}{2} \right)+11=0$

$3\left( k-1 \right)=15$

$3k-3=15$

$3k=15+3$

$3k=18$

$k=\dfrac{18}{3}$

$k=6$

8. (i) The point (-3, 2) lies on the line ax + 3y + 6 = 0, calculate the value of a. 

Ans: Given equation of the line is, ax + 3y + 6 = 0

Now substitute x = -3 and y = 2 in the given equation,

$a\left( -3 \right)+3\left( 2 \right)+6=0$

$-3a+6+6=0$

$-3a= -12$

$a=\dfrac{12}{3}$

$a=4$

(ii) The line y = mx + 8 contains the point (-4, 4), calculate the value of m.

Ans: Given equation of the line is, y = mx + 8

Substitute x = -4 and y = 4 in the given equation,

$4=m\left( -4 \right)+8$

$4-8= -4m$

$-4= -4m$

Therefore $m=1$

9. The point P divides the join of (2, 1) and (-3, 6) in the ratio 2 : 3. Does P lie on the line x - 5y + 15 = 0? 

Ans: Given equation of the line is, x - 5y + 15 = 0

In the ratio of 2: 3, the coordinate of the point that divides the join of (2, 1) and (-3, 6) will be,

$\left( x,y \right)=\left( \dfrac{m{{x}_{1}} + n{{y}_{1}}}{m + n},\dfrac{m{{x}_{2}} + n{{y}_{2}}}{m + n} \right)$

$\left( x,y \right)=\left( \dfrac{2 \times  \left( -3 \right) + 3 \times  2}{2 + 3},\dfrac{2 \times  6 + 3 \times  1}{2 + 3} \right)$

$\left( x,y \right)=\left( \dfrac{-6 + 6}{5},\dfrac{12 + 3}{5} \right)$

$\left( x,y \right)=\left( 0,\dfrac{15}{5} \right)$

$\left( x,y \right)=\left( 0,3 \right)$

Now, substitute x = 0 and y = 3 in the equation x - 5y + 15 = 0,

$0-5\left( 3 \right)+15=0$

$15+15=0$

$0=0$

Hence, the given statement is true and point P lies on the line.

10. The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio 1 : 2. Does the line x - 2y = 0 contain Q?

Ans: Given equation of the line x - 2y = 0

The point that divides the join of (5, -4) and (2, 2) in the ratio of 1 : 2 will have the coordinate,

$\left( x,y \right)=\left( \dfrac{m{{x}_{1}} + n{{y}_{1}}}{m + n},\dfrac{m{{x}_{2}} + n{{y}_{2}}}{m + n} \right)$

$\left( x,y \right)=\left( \dfrac{2 \times  5 + 1 \times  2}{2 + 1},\dfrac{2 \times  \left( -4 \right) + 1 \times  2}{2 + 1} \right)$

$\left( x,y \right)=\left( \dfrac{10 + 2}{3},\dfrac{-8 + 2}{3} \right)$

$\left( x,y \right)=\left( \dfrac{12}{3},\dfrac{-6}{3} \right)$

$\left( x,y \right)=\left( 4, -2 \right)$

Now substitute x = 4 and y = -2 in the given equation,

(4) - 2(-2) = 0

4 + 4 = 0

Hence, the given statement is not true as the line does not contain point Q.

11. Find the point of intersection of the line, 4x + 3y = 1 and 3x - y + 9 = 0. If this point lies on the line (2k - 1)x - 2y = 4: find the value of k. 

The above question can also be stated as:

If the lines 4x + 3y = 1, 3x - y + 9 = 0, and (2k - 1) x - 2y = 4 are concurrent (pass through the same point), find the value of k.

Ans: Equations of the line is 4x + 3y = 1 ------- (1)

And 3x - y + 9 = 0 ------- (2) $\times 3$

$\Rightarrow $ 9x - 3y + 27 = 0 ------- (3)

Now by adding equation (1) and (3), we get

$4x+3y+9x-3y=1-27$

$13x= -26$

$x=\dfrac{-26}{13}$

$x= -2$

Substitute the value of x in equation (1).

$4\left( -2 \right)+3y=1$

$3y=1+8$

$3y=9$

$y=\dfrac{9}{3}$

$y=3$

Given equation of other line is $\left( 2k-1 \right)x-2y=4$

Substitute x = -2 and y = 3 in the above equation,

$\left( 2k-1 \right)\left( -2 \right)-2\left( 3 \right)=4$

$-4k+2-6=4$

$-4k=4+4$

$-4k=8$

$k= -\dfrac{8}{4}$

$k= -2$

12. Show that the lines 2x + 5y = 1, x - 3y = 6 and x + 5y + 2 = 0 are concurrent.

Ans: Given equation of lines are,

2x + 5y = 1 ------- (1)

x - 3y = 6 ------- (2)

x + 5y + 2 = 0 ------- (3)

Subtracting equation (2) from equation (3), we get,

$x-3y-6-x-5y-2=0$

$-8y-8=0$

$y=\dfrac{-8}{8}$

$y= -1$

Substitute the value of y in equation (2),

$x-3\left( -1 \right)=6$

$x+3=6$

$x=6-3$

$x=3$

The intersection of lines 2 and 3 should be on the first line, making the lines concurrent.

Substitute x = 3 and y = -1 in equation (1),

$2\left( 3 \right)+5\left( -1 \right)=1$

$6-5=1$

$6-5=1$

$1=1$

Thus, the statement is true.

Hence the given lines are concurrent.

EXERCISE 14 (B)

1. Find the slope of the line whose inclination is: 

(i) 0° 

Ans: Slope of the line $\theta $= 0° will be,

$m=tan \theta $

$m=tan 0{}^\circ $

$m=0$

(ii) 30° 

Ans: Slope of the line $\theta $= 30° will be,

$m=tan \theta $

$m=tan 30{}^\circ $

$m=\dfrac{1}{\sqrt{3}}$

(iii) 72° 30' 

Ans: $\theta =72{}^\circ  30'$

$\theta =\left( 72\times \dfrac{30}{60} \right)$

$\theta =\left( 72\times \dfrac{1}{2} \right)$

$\theta =\left( 72\times 0.5 \right)$

Slope of the line $\theta $= 72.5° will be,

$m=tan 72.5{}^\circ $

$m=317159$

(iv) 46° 

Ans: Slope of the line $\theta $= 46° will be,

$m=tan \theta $

$m=tan 46{}^\circ $

$m=1.0355$

2. Find the inclination of the line whose slope is:

(i) 0 

Ans: The slope of the line is m = 0.

Therefore the inclination of the line will be,

$tan \theta =0$

$\theta =ta{{n}^{-1}} 0$

$\theta =0{}^\circ $

(ii) $\sqrt{3}$

Ans: The slope of the line is m =$\sqrt{3}$.

Therefore the inclination of the line will be,

$tan \theta =\sqrt{3}$

$\theta =ta{{n}^{-1}} \sqrt{3}$

$\theta =60{}^\circ $

(iii) 0.7646 

Ans: The slope of the line is m = 0.7646

Therefore the inclination of the line will be,

$tan \theta =0.7646$

$\theta =ta{{n}^{-1}} 0.7646$

$\theta =37.4{}^\circ $

$\theta =37{}^\circ  \left( 0.4\times 60 \right)'$

$\theta =37{}^\circ  24'$

(iv) 1.0875 

Ans: The slope of the line is m = 1.0875

Therefore the inclination of the line will be,

$tan \theta =1.0875$

$\theta =ta{{n}^{-1}} 1.0875$

$\theta =47.4{}^\circ $

$\theta =47{}^\circ  \left( 0.4\times 60 \right)'$

$\theta =47{}^\circ  24'$

3. Find the slope of the line passing through the following pairs of points: 

(i) (-2, -3) and (1, 2) 

Ans: Let (-2, -3) is $\left( {{x}_{1}},{{y}_{1}} \right)$ and (1, 2) is $\left( {{x}_{2}},{{y}_{2}} \right)$

Therefore the slope of line will be,

$m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{2 - \left( -3 \right)}{1 - \left( -2 \right)}$

$m=\dfrac{5}{3}$

(ii) (-4, 0) and origin 

Ans: Let (-4, 0) is $\left( {{x}_{1}},{{y}_{1}} \right)$ and (0, 0) is $\left( {{x}_{2}},{{y}_{2}} \right)$

Therefore the slope of line will be,

$m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{0 - \left( 0 \right)}{0 - \left( -4 \right)}$

$m=0$

(iii) (a, -b) and (b, -a)

Ans: Let (a, -b) is $\left( {{x}_{1}},{{y}_{1}} \right)$ and (b, -a) is $\left( {{x}_{2}},{{y}_{2}} \right)$

Therefore the slope of line will be,

$m=\dfrac{- a - \left( -b \right)}{b - a}$

$m=\dfrac{b - a}{b - a}$

$m=1$

4. Find the slope of the line parallel to AB if: 

(i) A = (-2, 4) and B = (0, 6) 

Ans: The slope of the line parallel to AB will be equal to the slope of AB.

$m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{6 - 4}{0 - \left( -2 \right)}$

$m=\dfrac{2}{2}$

$m=1$

(ii) A = (0, -3) and B = (-2, 5)

Ans: The slope of the line parallel to AB will be equal to the slope of AB.

$m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{5 - \left( -3 \right)}{-2 - 0}$

$m=\dfrac{8}{-2}$

$m= -4$

5. Find the slope of the line perpendicular to AB if: 

(i) A = (0, -5) and B = (-2, 4) 

Ans: Slope of line AB is,

$m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{4 - \left( -5 \right)}{-2 - 0}$

$m= -\dfrac{9}{2}$

Now, slope of line perpendicular to AB will be,

${{m}_{2}}= -\dfrac{1}{{{m}_{1}}}$

${{m}_{2}}= -\dfrac{1}{-\dfrac{9}{2}}$

${{m}_{2}}=\dfrac{2}{9}$

(ii) A = (3, -2) and B = (-1, 2)

Ans: Slope of line AB is,

$m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{2 - \left( -2 \right)}{-1 - 3}$

$m= -\dfrac{4}{4}$

$m= -1$

Now, slope of line perpendicular to AB will be,

${{m}_{2}}= -\dfrac{1}{{{m}_{1}}}$

${{m}_{2}}= -\dfrac{1}{-1}$

${{m}_{2}}=1$

6. The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.

Ans: W.K.T, slope of the line is, $m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

So, the slope of line passes through (0, 2) and (-3, -1) will be,

${{m}_{1}}=\dfrac{- 1 - 2}{- 3 - 0}$

${{m}_{1}}=\dfrac{- 3}{- 3}$

${{m}_{1}}=1$

Now, the slope of line passes through (-1, 5) and (4, a) will be,

${{m}_{2}}=\dfrac{a - 5}{4 - \left( -1 \right)}$

${{m}_{2}}=\dfrac{a - 5}{5}$

Since, both the lines are parallel. As a result the slope will be equal.

${{m}_{1}}={{m}_{2}}$

$1=\dfrac{a - 5}{5}$

$5=a-5$

$a=10$

7. The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.

Ans: W.K.T, slope of the line is, $m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

The slope of line passes through (-4, -2) and (2, -3) will b,

${{m}_{1}}=\dfrac{- 3 - \left( - 2 \right)}{2 - \left( - 4 \right)}$

${{m}_{1}}=\dfrac{- 1}{6}$

Now, the slope of line passes through (a, 5) and (2, -1) will be,

${{m}_{2}}=\dfrac{- 1 - 5}{2 - a}$

${{m}_{2}}=\dfrac{- 6}{2 - a}$

Since, both the lines are perpendicular.

${{m}_{1}}{{m}_{2}}= -1$

$\dfrac{- 1}{6}\times \dfrac{- 6}{2 - a}= -1$

$\dfrac{1}{2 - a}= -1$

$-2+a=1$

$a=3$

8. Without using the distance formula, show that the points A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.

Ans: Given points are A (4, -2), B (-4, 4) and C (10, 6).

W.K.T, slope of the line is, $m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

So, the slope of line AB will be,

${{m}_{1}}=\dfrac{4 - \left( - 2 \right)}{- 4 - 4}$

${{m}_{1}}=\dfrac{6}{- 8}$

${{m}_{1}}= -\dfrac{3}{4}$

Slope of line BC will be,

${{m}_{2}}=\dfrac{6 - 4}{10 - \left( - 4 \right)}$

${{m}_{2}}=\dfrac{2}{14}$

${{m}_{2}}=\dfrac{1}{7}$

Now, slope of line AC will be,

${{m}_{3}}=\dfrac{6 - \left( - 2 \right)}{10 - 4}$

${{m}_{3}}=\dfrac{8}{6}$

${{m}_{3}}=\dfrac{4}{3}$

Here, 

${{m}_{1}}{{m}_{3}}=\dfrac{-3}{4}\times \dfrac{4}{3}$

${{m}_{1}}{{m}_{3}}= -1$

Thus, AB is perpendicular to AC.

Therefore, the given points are the vertices of the right angle triangle.

9. Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of parallelogram.

Ans: Given points are A (4, 5), B (1, 2), C (4, 3) and D (7, 6)

W.K.T, slope of the line is, $m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

So, the slope of line AB will be,

${{m}_{1}}=\dfrac{2 - 5}{1 - 4}$

${{m}_{1}}=\dfrac{- 3}{- 3}$

${{m}_{1}}=1$

Slope of line BC will be,

${{m}_{2}}=\dfrac{3 - 2}{4 - 1}$

${{m}_{2}}=\dfrac{1}{3}$

Slope of line CD will be,

${{m}_{3}}=\dfrac{6 - 3}{7 - 4}$

${{m}_{3}}=\dfrac{3}{3}$

${{m}_{3}}=1$

Slope of line AD will be,

${{m}_{4}}=\dfrac{6 - 5}{7 - 4}$

${{m}_{4}}=\dfrac{1}{3}$

Here, ${{m}_{1}}={{m}_{3}}$ and ${{m}_{2}}={{m}_{4}}$

Therefore line AB | | CD and line BC | | AD.

Hence the given points are vertices of the parallelogram.

10. (-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the midpoints of its sides, is a parallelogram.

Ans: Given points are A = (-2, 4), B = (4, 8), C = (10, 7) and D = (11, -5)

W.K.T coordinate of the midpoint formula is, $P=\left( \dfrac{{{x}_{2}} + {{x}_{1}}}{2},\dfrac{{{y}_{2}} + {{y}_{1}}}{2} \right)$

The coordinate of the midpoint of AB will be,

$P=\left( \dfrac{4 + \left( -2 \right)}{2},\dfrac{8 + 4}{2} \right)$

$P=\left( \dfrac{2}{2},\dfrac{12}{2} \right)$

$P=\left( 1, 6 \right)$

The coordinate of the midpoint of BC will be,

$Q=\left( \dfrac{4 + 10}{2},\dfrac{8 + 7}{2} \right)$

$Q=\left( \dfrac{14}{2},\dfrac{15}{2} \right)$

$Q=\left( 7, 7.5 \right)$

The coordinate of the midpoint of CD will be,

$R=\left( \dfrac{11 + 10}{2},\dfrac{- 5 + 7}{2} \right)$

$R=\left( \dfrac{21}{2},\dfrac{2}{2} \right)$

$R=\left( 10.5, 1 \right)$

The coordinate of the midpoint of AD will be,

$S=\left( \dfrac{11 + \left( - 2 \right)}{2},\dfrac{- 5 + 4}{2} \right)$

$S=\left( \dfrac{9}{2},\dfrac{-1}{2} \right)$

$S=\left( 4.5, -0.5 \right)$

W.K.T slope of the line is $m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

Now, slope of line PQ will be,

${{m}_{1}}=\dfrac{7.5 - 6}{7 - 1}$

${{m}_{1}}=\dfrac{1.5}{6}$

${{m}_{1}}=\dfrac{15}{60}$

${{m}_{1}}=\dfrac{1}{4}$

Slope of line QR will be,

${{m}_{2}}=\dfrac{1 - 7.5}{10.5 - 7}$

${{m}_{2}}=\dfrac{- 6.5}{3.5}$

${{m}_{2}}=\dfrac{-13}{7}$

Slope of line RS will be,

${{m}_{3}}=\dfrac{0.5 - 1}{4.5 - 10.5}$

${{m}_{3}}=\dfrac{- 1.5}{- 6}$

${{m}_{3}}=\dfrac{-15}{- 60}$

${{m}_{3}}=\dfrac{1}{4}$

Slope of line PS will be,

${{m}_{4}}=\dfrac{- 0.5 - 6}{4.5 - 1}$

${{m}_{4}}=\dfrac{- 6.5}{3.5}$

${{m}_{4}}=\dfrac{- 65}{35}$

${{m}_{4}}=\dfrac{- 13}{7}$

Here, ${{m}_{1}}={{m}_{3}}$ and ${{m}_{2}}={{m}_{4}}$

Therefore line PQ | | RS and line QR | | PS.

Hence the given points are vertices of the parallelogram.

11. Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are collinear.

Ans: The given points are P = (a, b + c), Q = (b, c + a) and R = (c, a + b)

W.K.T, slope of the line is, $m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

Slope of line PQ will be,

${{m}_{1}}=\dfrac{\left( c + a \right) - \left( b + c \right)}{b - a}$

${{m}_{1}}=\dfrac{c + a - b - c}{b - a}$

${{m}_{1}}=\dfrac{a - b}{b - a}$

${{m}_{1}}=\dfrac{a - b}{-\left( a - b \right)}$

${{m}_{1}}= -1$

Slope of line QR will be,

${{m}_{2}}=\dfrac{\left( a + b \right) - \left( c + a \right)}{c - b}$

${{m}_{2}}=\dfrac{a + b - c - a}{c - b}$

${{m}_{2}}=\dfrac{b - c}{c - b}$

${{m}_{2}}=\dfrac{b - c}{-\left( b - c \right)}$

${{m}_{2}}= -1$

Here, ${{m}_{1}}={{m}_{2}}$

Therefore, the points are collinear

Hence proved.

12. Find x, if the slope of the line joining (x, 2) and (8, -11) is $-\dfrac{3}{4}$.

Ans: Let A = (x, 2) and B = (8, -11)

Therefore slope of the line will be, $m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$-\dfrac{3}{4}=\dfrac{-11 - 2}{8 - x}$

$-3\left( 8-x \right)=4\left( -11-2 \right)$

$-24+3x= -44-8$

$3x= -52+24$

$3x= -28$

$x=\dfrac{-28}{3}$

13. The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slopes of all its sides.

${{m}_{AC}}=tan 60$

${{m}_{AC}}=\sqrt{3}$

Ans: The side AB is perpendicular to the x axis.

W.K.T, slope of the side is, $m=tan \theta $

As a result, slope of side AB will be ,

${{m}_{AB}}=tan 0$

${{m}_{AB}}=0$

In an equilateral triangle all the angles are 60°.

The slope of side AC will be,

${{m}_{AC}}=tan 60$

${{m}_{AC}}=\sqrt{3}$

And, the slope of side BC will be,

${{m}_{AC}}=tan 120$

${{m}_{AC}}= -\sqrt{3}$

14. The side AB of a square ABCD is parallel to the x-axis. Find the slopes of all its sides. Also, find: 

(i) the slope of the diagonal AC, 

(ii) the slope of the diagonal BD.

Ans: 

Side AB is parallel to the x - axis. 

Therefore the slope of side AB will be,

${{m}_{AB}}=tan 0$

${{m}_{AB}}=0$

As CD | | AB, slope of CD = slope of AB

Therefore ${{m}_{AB}}={{m}_{CD}}=0$

As AD$\bot $AB, slope of side AD will be

${{m}_{AD}}=tan 90$

${{m}_{AD}}=\infty $ or not defined

As BC | | AD, Therefore the slope of side BC will be,

${{m}_{BC}}={{m}_{AD}}=\infty $ or not defined

(i) With the positive x-axis direction, diagonal AC forms a ${{45}^{o}}$ angle.

Slope of AC = tan${{45}^{o}}$= 1

(ii) With the positive x-axis direction, diagonal BD forms a $-{{45}^{o}}$ angle.

Slope of AC = tan$\left( -{{45}^{o}} \right)$= -1

15. A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find: 

(i) the slope of the altitude of AB, 

(ii) the slope of the median AD and 

(iii) the slope of the line parallel to AC.

Ans: The given points are A (5, 4), B (-3, -2) and C (1, -8)

(i) Slope of AB,

${{m}_{AB}}=\dfrac{- 2 - 4}{- 3 - 5}$

${{m}_{AB}}=\dfrac{- 6}{- 8}$

${{m}_{AB}}=\dfrac{3}{4}$

Now, the slope of the altitude of AB = $-\dfrac{1}{Slope of AB}= -\dfrac{1}{\dfrac{3}{4}}= -\dfrac{4}{3}$

(ii) Since D is the midpoint of BC, the coordinate of point D will be,

$D=\left( \dfrac{- 3 + 1}{2},\dfrac{- 2 - 8}{2} \right)$

$D=\left( \dfrac{- 2}{2},\dfrac{- 10}{2} \right)$

$D=\left( -1,-5 \right)$

Therefore the slope of AD will be,

${{m}_{AD}}=\dfrac{- 5 - 4}{- 1 - 5}$

${{m}_{AD}}=\dfrac{- 9}{- 6}$

${{m}_{AD}}=\dfrac{3}{2}$

(iii) The slope of line AC will be,

${{m}_{AC}}=\dfrac{- 8 - 4}{1 - 5}$

${{m}_{AC}}=\dfrac{- 12}{- 4}$

${{m}_{AC}}=3$

Therefore the slope of line parallel to AC = Slope of line AC = 3

16. The slope of the side BC of a rectangle ABCD is $\dfrac{2}{3}$ Find:

(i) the slope of the side AB.

(ii) the slope of the side AD.

Ans: 

(i) Given that the slope of the side BC is $\dfrac{2}{3}$

BC is perpendicular to AB,

${{m}_{AB }}{{m}_{BC}}= -1$

${{m}_{AB }}=\dfrac{-1}{{{m}_{BC}}}$

${{m}_{AB }}=\dfrac{-1}{\dfrac{2}{3}}$

${{m}_{AB }}=\dfrac{- 3}{2}$

(ii) Since AD is parallel to BC

Slope of AD = Slope of BC = $\dfrac{2}{3}$

17. Find the slope and the inclination of the line AB if: 

(i) A = (-3, -2) and B = (1, 2) 

Ans: Given A = (-3, -2) and B = (1, 2)

Then, slope of AB is,

$m=\dfrac{2 - \left( -2 \right)}{1 - \left( -3 \right)}$

$m=\dfrac{4}{4}$

$m=1$

Now, the inclination of line AB will be,

$tan \theta =m$

$tan \theta =1$

$\theta =ta{{n}^{-1}}\left( 1 \right)$

$\theta ={{45}^{o}}$

(ii) A = (0, $\sqrt{3}$) and B = (3, 0) 

Ans: Given A = (0, $\sqrt{3}$) and B = (3, 0)

Then, slope of AB is,

$m=\dfrac{0 - \left( -\sqrt{3} \right)}{3 - 0}$

$m=\dfrac{\sqrt{3}}{3}$

$m=\dfrac{1}{\sqrt{3}}$

Now, the inclination of line AB will be,

$tan \theta =m$

$tan \theta =\dfrac{1}{\sqrt{3}}$

$\theta =ta{{n}^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$

$\theta ={{30}^{o}}$

(iii) A = (-1, 2$\sqrt{3}$) and B = (-2, $\sqrt{3}$)

Ans: Given A = (-1, 2$\sqrt{3}$) and B = (-2, $\sqrt{3}$)

Then, slope of AB is,

$m=\dfrac{\sqrt{3} - 2\sqrt{3})}{-2 - \left( -1 \right)}$

$m=\dfrac{-\sqrt{3}}{-1}$

$m=\sqrt{3}$

Now, the inclination of line AB will be,

$tan \theta =m$

$tan \theta =\sqrt{3}$

$\theta =ta{{n}^{-1}}\left( \sqrt{3} \right)$

$\theta ={{60}^{o}}$

18. The points (-3, 2), (2, -1) and (a, 4) are collinear. Find a.

Ans: Given points are A = (-3, 2), B = (2, -1) and C = (a, 4)

Since the points are collinear so,

Slope of AB = Slope of BC

${{m}_{AB}}={{m}_{BC}}$

$\dfrac{- 1 - 2}{2 + 3}=\dfrac{4 + 1}{a - 2}$

$\dfrac{- 3}{5}=\dfrac{5}{a - 2}$

$-3\left( a-2 \right)=25$

$-3a+6=25$

$-3a=19$

$a=\dfrac{19}{-3}$

$a= -6\dfrac{1}{3}$

19. The points (K, 3), (2, -4) and (-K + 1, -2) are collinear. Find K. 

Ans: Given A = (K, 3), B = (2, -4) and C = (-K + 1, -2)

Since the points are collinear so,

Slope of AB = Slope of BC

${{m}_{AB}}={{m}_{BC}}$

$\dfrac{- 4 - 3}{2 - K}=\dfrac{- 2 + 4}{- K + 1 - 2}$

$\dfrac{- 7}{2 - K}=\dfrac{2}{- K - 1}$

$-7\left( -K-1 \right)=2\left( 2-K \right)$

$7K+7=4-2K$

$7K+2K=4-7$

$9K= -3$

$K= \dfrac{-3}{9}$

$K= \dfrac{-1}{3}$

20. Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C. Which segment appears to have the steeper slope, AB or AC? Justify your conclusion by calculating the slopes of AB and AC.

Ans: Given A = (1, 1), B = (4, 7) and C = (4, 10)

From the graph it is clear that the segment AC is steeper than segment AB.

Slope of line AB,

${{m}_{AB}}=\dfrac{7 - 1}{4 - 1}$

${{m}_{AB}}=\dfrac{6}{3}$

${{m}_{AB}}=2$

And the slope of line AC,

${{m}_{AC}}=\dfrac{10 - 1}{4 - 1}$

${{m}_{AC}}=\dfrac{9}{3}$

${{m}_{AC}}=3$

Hence, the slope of AC is more than the slope of AB.

Therefore AC is steeper than AB.

21. Find the value(s) of k so that PQ will be parallel to RS. Given: 

(i) P(2, 4), Q(3, 6), R(8, 1) and S(10, k) 

Ans: Given points are P = (2, 4), Q = (3, 6), R = (8, 1) and S = (10, k)

Since PQ | | RS

Slope of PQ = Slope of RS

$\dfrac{6 - 4}{3 - 2}=\dfrac{k - 1}{10 - 8}$

$2=\dfrac{k - 1}{2}$

$4=k-1$

$k=5$

(ii) P(3, -1), Q(7, 11), R(-1, -1) and S(1, k) 

Ans: Given points are P = (3, -1), Q = (7, 11), R = (-1, -1) and S = (1, k)

Since PQ | | RS

Slope of PQ = Slope of RS

$\dfrac{11-(-1)}{7-3} = \dfrac{k-(-1)}{1-(-1)}$

$\dfrac{12}{4} = \dfrac{k+1}{2}$

$\dfrac{k+1}{2}=3$

$k+1=6$

$k=5$

(iii) P(5, -1), Q(6, 11), R(6, -4k) and S(7, ${{k}^{2}}$) 

Ans: Given points are P = (3, -1), Q = (7, 11), R = (-1, -1) and S = (1, k) 

Since PQ | | RS

Slope of PQ = Slope of RS

$\dfrac{11 - \left( -1 \right)}{6 - 5}=\dfrac{{{k}^{2}} - \left( -4k \right)}{7 - 6}$

$12={{k}^{2}} + 4k$

${{k}^{2}}+4k-12=0$

On factorising,

${{k}^{2}}+6k-2k-12=0$

$k\left( k+6 \right)-2\left( k+6 \right)=0$

$\left( k+6 \right)\left( k-2 \right)=0$

On equating to zero we get,

k + 6 = 0 or k - 2 = 0

k = -6 or k = 2

Therefore the value of k is -6 or 2

EXERCISE 14 (C)

1. Find the equation of a line whose: 

y-intercept = 2 and slope = 3

Ans: Given, y-intercept (c) = 2 and slope ( m ) = 3.

The equation of the line is, y = mx + c

On substituting the values i.e., m = 3 and c = 2 in the above equation. 

We get, 

y = 3x + 2 as the required equation.

2. Find the equation of a line whose: 

y-intercept = -1 and inclination = 45°

Ans: Given y-intercept ( c ) = -1 and inclination = 45°

Therefore slope ( m ) is,

$m=tan \theta $

$m=tan {{45}^{o}}$

$m=1$

Now substitute the value of c = -1 and m = 1 in the equation of the line y = mx + c,

We get, y = 1(x) + (-1)

Therefore y = x - 1 is the required equation.

3. Find the equation of the line whose slope is $-\dfrac{4}{3}$ which passes through (-3, 4).

Ans: Given slope (m) = $-\dfrac{4}{3}$ 

Line passes through the point (-3, 4) = $\left( {{x}_{1}},{{y}_{1}} \right)$

Thus the equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-4= -\dfrac{4}{3} \left( x-\left( -3 \right) \right)$

$3\left( y-4 \right)= -4 \left( x+3 \right)$

$3y-12= -4x-12$

$3y= -4x$

$3y+4x=0$

Hence the required equation.

4. Find the equation of a line which passes through (5, 4) and makes an angle of 60° with the positive direction of the x-axis.

Ans: Given, the line passes through the point $({{x}_{1}},{{y}_{1}})$ = (5, 4) 

Angle between the x-axis and the line is 60°.

Therefore slope of the line is,

$m=tan \theta $

$m=tan {{60}^{o}}$

$m=\sqrt{3}$

On substituting the values in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$, we get

$y-4=\sqrt{3}\left( x-5 \right)$

$y=\sqrt{3}x-5\sqrt{3}+4$

5. Find the equation of the line passing through: 

(i) (0, 1) and (1, 2) 

Ans: Given points are $\left( {{x}_{1}},{{y}_{1}} \right)$ = (0, 1) and $\left( {{x}_{2}},{{y}_{2}} \right)$ = (1, 2).

Slope of the line, $m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{2 - 1}{1 - 0}$

$m=1$

On substituting the values in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$, we get

$y-1=1\left( x-0 \right)$

$y=x+1$

(ii) (-1, -4) and (3, 0) 

Ans: Given points are $\left( {{x}_{1}},{{y}_{1}} \right)$ = (-1, -4) and $\left( {{x}_{2}},{{y}_{2}} \right)$ = (3, 0).

Slope of the line, $m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{0 - \left( -4 \right)}{3 - \left( -1 \right)}$

$m=\dfrac{4}{4}$

$m=1$

On substituting the values in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$, we get

$y-\left( -4 \right)=1\left( x-\left( -1 \right) \right)$

$y+4=x+1$

$y=x-3$

6. The coordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find: 

(i) the gradient of PQ; 

(ii) the equation of PQ; 

(iii) the coordinates of the point where PQ intersects the x-axis.

Ans: Given points are P = (2, 6) and Q = (-3, 5)

(i) The gradient of PQ is,

$m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{5 - 6}{- 3 - 2}$

$m=\dfrac{- 1}{- 5}$

$m=\dfrac{1}{5}$

(ii) The equation of the line PQ is,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-6=\dfrac{1}{5}\left( x-2 \right)$

$5\left( y-6 \right)=\left( x-2 \right)$

$5y-30=x-2$

$5y=x+30-2$

$5y=x+28$

$y=\dfrac{1}{5}x+\dfrac{28}{5}$

The above equation is in the form y = mx + c, where c = $\dfrac{28}{5}$

(iii) The coordinate of point PQ intersect the x-axis at point (x, 0)

Now, substitute y = 0 in the equation of the line, y = mx + c

$0=\dfrac{1}{5}x+\dfrac{28}{5}$

Taking LCM and equating we get,

$x+28=0$

$x= -28$

Therefore the coordinate will be (-28, 0)

7. The coordinates of two points A and B are (-3, 4) and (2, -1). Find: 

(i) the equation of AB; 

(ii) the coordinates of the point where the line AB intersects the y-axis. 

Ans: Given points are A = (-3, 4) and B = (2, -1).

(i) The slope of the line is, $m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{-1 - 4}{2 - \left( -3 \right)}$

$m=\dfrac{-5}{5}$

$m= -1$

Equation of the line AB is given by:

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-4= -1\left( x-\left( -3 \right) \right)$

$y-4= -x-3$

$y=1-x$

$y+x=1$

The equation is the form y = mx + c, where c = 1

(ii) The coordinate of point AB intersect the y-axis at point (0, y)

Substitute x = 0 in the equation of the line

y + 0 = 1

y = 1

Therefore the coordinates are (0, 1).

8. The figure given alongside shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equations of AB and CD. 

Ans: The point of intersection is $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 3,4 \right)$

Slope of line AB = m = $tan {{45}^{o}}$ = 1

Equation of line AB is,

$y-4=1\left( x-3 \right)$

$y=x-3+4$

$y=x+1$

Slope of line CD = m = $tan {{45}^{o}}=\sqrt{3}$

The equation of line CD is, 

$y-4=\sqrt{3}\left( x-3 \right)$

$y=\sqrt{3}x-3\sqrt{3}+4$

9. In $\Delta $ABC, A = (3, 5), B = (7, 8) and C = (1, -10). Find the equation of the median through A.

Ans: The given points are A = (3, 5), B = (7, 8) and C = (1, -10).

Here D is the midpoint of BC, therefore the coordinates of D is,

$D=\left( \dfrac{7 + 1}{2},\dfrac{8 + \left( -10 \right)}{2} \right)$

$D=\left( \dfrac{8}{2},\dfrac{-2}{2} \right)$

$D=\left( 4, -1 \right)$

Slope of AD $=m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{-1 - 5}{4 - 3}$

$m= -6$

Therefore the equation of median AD is,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-5= -6\left( x-3 \right)$

$y-5= -6x+18$

$y+6x=23$

10. The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, $\hat{A}={{60}^{o}}$ and vertex C = (7, 5). Find the equations of BC and CD.

Ans: Given that ABCD is a parallelogram and side AB is parallel to the x-axis.

Therefore CD | | AB and BC | | AD.

$\hat{A}=\hat{B}={{180}^{o}}$

So, $\hat{B}={{180}^{o}}-{{60}^{o}}={{120}^{o}}$  (Because $\hat{A}={{60}^{o}}$)

Slope of side BC, $m=tan {{120}^{o}}=tan\left( {{90}^{o}}+{{30}^{o}} \right)=cot {{30}^{o}}=\sqrt{3}$

Equation of BC will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-5=\sqrt{3}\left( x-7 \right)$

$y=\sqrt{3}x-7\sqrt{3}+5$

As we know that CD | | AB and AB are parallel to the x-axis.

Slope of CD = Slope of AB = 0.

Equation of the line CD is,

$y-5=0\left( x-7 \right)$

$y=5$

11. Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x - y = 4. 

Ans: The given equation of the lines are, 

x + 2y = 7  --------(1)

And, x - y = 4  --------(2)

Subtract equation (2) from (1);

x - y - x - 2y = 4 - 7

-3y = -3

y = 1

Now, substitute y value in equation (1)

x + 2(1) = 7

x = 5

Hence, the coordinate of the intersection is (5, 1)

The slope of the line passes through the origin (0, 0) and intersection point (5, 1) will be,

$m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{1 - 0}{5 - 0}$

$m=\dfrac{1}{5}$

Thus the required equation of the line is,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-0=\dfrac{1}{5}\left( x-0 \right)$

$5y=x$

Therefore, $x-5y=0$.

12. In triangle ABC, the coordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A. 

Also, find the equation of the line through vertex B and parallel to AC.

Ans: Given, coordinates of the triangle ABC are A = (4, 7), B = (-2, 3) and C = (0, 1)

The median through vertex A bisects the side BC, So D is the midpoint on side BC.

Therefore the coordinate of point D will be,

$D=\left( \dfrac{{{x}_{2}} - {{y}_{2}}}{2},\dfrac{{{x}_{1}} - {{y}_{1}}}{2} \right)$

$D=\left( \dfrac{-2 - 0}{2},\dfrac{3 + 1}{2} \right)$

$D=\left( \dfrac{-2}{2},\dfrac{4}{2} \right)$

$D=\left( -1,2 \right)$           

Slope of AD is,

${{m}_{AD}}=\dfrac{2 - 7}{-1 - 4}$

${{m}_{AD}}=\dfrac{-5}{-5}$

${{m}_{AD}}=1$

Therefore the equation of median AD is given by,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-7=1\left( x-4 \right)$

$y=x-4+7$

$y=x+3$

Slope of AC will be,

${{m}_{AC}}=\dfrac{1 - 7}{0 - 4}$

${{m}_{AC}}=\dfrac{-6}{-4}$

${{m}_{AC}}=\dfrac{3}{2}$

Therefore the equation of the line parallel to AC and passing through vertex B will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-3=\dfrac{3}{2}\left( x-\left( -2 \right) \right)$

$2\left( y-3 \right)=3\left( x+2 \right)$

$2y-6=3x+6$

$2y=3x+12$

13. A, B and C have coordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.

Ans: Given coordinates are A = (0, 3), B = (4, 4) and C = (8, 0)

Slope of BC is,

${{m}_{BC}}=\dfrac{0 - 4}{8 - 4}$

${{m}_{BC}}=\dfrac{-4}{4}$

${{m}_{BC}}= -1$

Slope of the line perpendicular to BC will be,

$m {{m}_{BC}}= -1$

$m\left( -1 \right)= -1$

$m=1$

Therefore the equation of line perpendicular to BC and passing through vertex A will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-3=1\left( x-0 \right)$

$y=x+3$

14. Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3). 

Ans: Let A = (1, 4), B = (2, 3) and C = (-1, 2)

Slope of AB,

${{m}_{AB}}=\dfrac{3 - 4}{2 - 1}$

${{m}_{AB}}= -1$

Now, slope of the line perpendicular to BC will be,

$m {{m}_{AB}}= -1$

$m \left( -1 \right)= -1$

$m=1$

Equation of line perpendicular to BC and passing through point A will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-2=1\left( x-\left( -1 \right) \right)$

$y=x+3$

15. Find equation of the line whose: 

(i) x-intercept = 5 and y-intercept = 3 

Ans: Given, x-intercept = 5 and y-intercept = 3.

Therefore the equation of the line will be,

$\dfrac{x}{a}+\dfrac{y}{b}=1$

$\dfrac{x}{5}+\dfrac{y}{3}=1$

Taking LCM, we get

$\dfrac{3x + 5y}{15}=1$

$3x+5y=15$

(ii) x-intercept = -4 and y-intercept = 6 

Ans: Given, x-intercept = -4 and y-intercept = 6.

Therefore the equation of the line will be,

$\dfrac{x}{a}+\dfrac{y}{b}=1$

$\dfrac{x}{-4}+\dfrac{y}{6}=1$

Taking LCM, we get

$\dfrac{3x - 2y}{-12}=1$

$3x-2y= -12$

(iii) x-intercept = -8 and y-intercept = -4 

Ans: Given, x-intercept = -8 and y-intercept = -4.

Therefore the equation of the line will be,

$\dfrac{x}{a}+\dfrac{y}{b}=1$

$\dfrac{x}{-8}+\dfrac{y}{-4}=1$

Taking LCM, we get

$\dfrac{x + 2y}{-8}=1$

$x+2y= -8$

16. Find the equation of the line whose slope is $\dfrac{-5}{6}$ and x-intercept is 6. 

Ans: Given, slope of the line is $\dfrac{-5}{6}$ and x-intercept is 6.

Therefore the coordinate on x-axis will be (6, 0).

The equation of the line is,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-0=\dfrac{-5}{6}\left( x-6 \right)$

$6y= -5\left( x-6 \right)$

$6y= -5x+30$

$5x+6y=30$

17. Find the equation of the line with x-intercept 5 and a point on it (-3, 2). 

Ans: Let the point is $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -3,2 \right)$ and the coordinate of the x-intercept is $({{x}_{2}},{{y}_{2}})=\left( 5,0 \right)$

Slope of the line will be,

$m=\dfrac{0 - 2}{5 - \left( -3 \right)}$

$m=\dfrac{-2}{8}$

$m=\dfrac{-1}{4}$

Equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-2=\dfrac{-1}{4}\left( x-\left( -3 \right) \right)$

$4\left( y-2 \right)= -1\left( x+3 \right)$

$4y-8= -x-3$

$x+4y=5$

18. Find the equation of the line through (1, 3) and make an intercept of 5 on the y-axis. 

Ans: Let the point is $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -3,2 \right)$ and the coordinate of the y-intercept is $({{x}_{2}},{{y}_{2}})=\left( 5,0 \right)$

Slope of the line will be,

$m=\dfrac{5 - 3}{0 - 1}$

$m= -2$

Equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-3= -2\left( x-1 \right)$

$y=3-2x+2$

$y+2x=5$

19. Find the equations of the lines passing through point (-2, 0) and equally inclined to the coordinate axes. 

Ans: The angle between the line and x-axis when the line is equally inclined with the coordinate axes will be 45°.

Therefore the possible slopes of the line will be,

${{m}_{1}}=tan {{45}^{o}}=1$

${{m}_{2}}=tan \left( -{{45}^{o}} \right)= -1$

Equation of the lines that passes through the point (-2, 0) will be,

For m = 1,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-0=1\left( x-\left( -2 \right) \right)$

$y=x+2$

For m = -1,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-0= -1\left( x-\left( -2 \right) \right)$

$y= -x-2$

$y+x+2=0$

20. The line through P(5, 3) intersects y-axis at Q. 

(i) Write the slope of the line. 

(ii) Write the equation of the line. 

(iii) Find the coordinates of Q. 

Ans: 

(i) The inclination of the line with the x-axis is 45°.

So the slope of the line will be, 

$m=tan {{45}^{o}}=1$

(ii) Given line is passing through the point (5, 3).

Therefore the equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-3=1\left( x-5 \right)$

$y=x-5+3$

$y=x-2$

(iii) Point Q is the intercept on the y-axis. Therefore the coordinate of point Q will be (0, y).

Now, substitute x = 0 in the equation of the line,

$y=x-2$

$y=0-2$

$y= -2$

Therefore the coordinate of the point Q will be (0, -2) 

21. Write down the equation of the line whose gradient is $-\dfrac{2}{5}$ and which passes through point P, where P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3 : 1.

Ans: Given, gradient of line is $-\dfrac{2}{5}$ and point P divides the line segment joining points A = (4, -8) and B = (12, 0) in the ratio 3 : 1.

Therefore the coordinate of point P will be,

$P=\left( \dfrac{m{{x}_{1}} + n{{x}_{2}}}{m + n},\dfrac{m{{y}_{1}} + n{{y}_{2}}}{m + n} \right)$

$P=\left( \dfrac{3 \times  12 + 1 \times  4}{3 + 1},\dfrac{3 \times  0 + 1 \times  \left( -8 \right)}{3 + 1} \right)$

$P=\left( \dfrac{36 + 4}{4},\dfrac{-8}{4} \right)$

$P=\left( \dfrac{40}{4},\dfrac{-8}{4} \right)$

$P=\left( 10, -2 \right)$

Equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-\left( -2 \right)= -\dfrac{2}{5}\left( x-10 \right)$

$5\left( y+2 \right)= -2\left( x-10 \right)$

$5y+10= -2x+20$

$5y+2x=20-10$

$2x+5y=10$

22. A (1, 4), B (3, 2) and C (7, 5) are vertices of a triangle ABC. Find: 

(i) the coordinates of the centroid of triangle ABC. 

(ii) the equation of a line, through the and parallel to AB. 

Ans: (i) Given vertices of the triangle are A = (1, 4), B = (3, 2) and C = (7, 5)

The coordinate of the centroid of the triangle ABC will be,

$P=\left( \dfrac{{{x}_{1}} + {{x}_{2}} + {{x}_{3}}}{3},\dfrac{{{y}_{1}} + {{y}_{2}} + {{y}_{3}}}{3} \right)$

$P=\left( \dfrac{1 + 3 + 7}{3},\dfrac{4 + 2 + 5}{3} \right)$

$P=\left( \dfrac{11}{3},\dfrac{11}{3} \right)$

(ii) Slope of AB will be,

${{m}_{AB}}=\dfrac{2 - 4}{3 - 1}$

${{m}_{AB}}=\dfrac{-2}{2}$

${{m}_{AB}}= -1$

Equation of line parallel to AB and passes through the centroid will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-\dfrac{11}{3}= -1\left( x-\dfrac{11}{3} \right)$

On taking LCM, we get

$\dfrac{3y - 11}{3}= \dfrac{-3x + 11}{3}$

$3y-11=11-3x$

$3y+3x=22$

23. A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point P in AC; such that AP : CP = 2 : 3. 

Ans: Given A = (7, -1), B = (4, 1) and C = (-3, 4)

Point P on side AC is such that AP : CP = 2 : 3.

Therefore the coordinate of point P will be,

$P=\left( \dfrac{m{{x}_{1}} + n{{x}_{2}}}{m + n},\dfrac{m{{y}_{1}} + n{{y}_{2}}}{m + n} \right)$

$P=\left( \dfrac{\left( -3 \right) \times  2 + 7 \times  3}{2 + 3},\dfrac{4 \times  2 + \left( -1 \right) \times  3}{2 + 3} \right)$

$P=\left( \dfrac{-6 + 21}{5},\dfrac{8 - 3}{5} \right)$

$P=\left( \dfrac{15}{5},\dfrac{5}{5} \right)$

$P=\left( 3,1 \right)$

The slope of the line that passes through points B = (4, 1) and P = (3, 1) will be,

$m=\dfrac{1 - 1}{3 - 4}$

$m=0$

Therefore the equation of line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-1=0\left( x-4 \right)$

$y-1=0$

$y=1$

EXERCISE 14 (D)

1. Find the slope and y-intercept of the line: 

(i) y = 4 

Ans: Given equation is, y = 4

On comparing with equation of the line, y = mx + c 

We get,

Slope is m = 0

y-intercept = 4.

(ii) ax - by = 0 

Ans: Given equation is, ax - by = 0

Therefore, $ax=by$

$y=\dfrac{a}{b}x$

On comparing with equation of the line, y = mx + c 

We get,

Slope is m = $\dfrac{a}{b}$

y-intercept = 0.

(iii) 3x - 4y = 5 

Ans: Given equation is, 3x - 4y = 5 

Therefore, $4y=3x-5$

$y=\dfrac{3}{4}x-\dfrac{5}{4}$

On comparing with equation of the line, y = mx + c 

We get,

Slope is m = $\dfrac{3}{4}$

y-intercept = $-\dfrac{5}{4}$

2. The equation of a line is x - y = 4. Find its slope and y-intercept. Also, find its inclination.

Ans: Given equation is, x - y = 4 

Therefore, $y=x-4$

On comparing with equation of the line, y = mx + c 

We get,

Slope is m = 1

y-intercept = -4

And, inclination will be,

$m=tan \theta $

$tan \theta =1$

$\theta =ta{{n}^{-1}}\left( 1 \right)$

$\theta ={{45}^{o}}$

3. (i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x - 21y + 50 = 0? 

Ans: Given lines are 3x + 4y + 7 = 0 and 28x - 21y + 50 = 0.

Now, $3x+4y+7=0$,

$4y= -3x-7$

$y= -\dfrac{3x}{4}-\dfrac{7}{4}$

Compare with, $y=mx+c$

$c= -\dfrac{7}{4}$  and  ${{m}_{1}}= -\dfrac{3}{4}$

Now for equation $28x-21y+50=0$,

$-21y= -28x-50$ -------$\times \left( -1 \right)$

$y=\dfrac{28x}{21}+\dfrac{50}{21}$

$y=\dfrac{4x}{3}+\dfrac{50}{21}$

Compare with, $y=mx+c$

$c=\dfrac{50}{21}$  and  ${{m}_{2}}=\dfrac{4}{3}$

Now consider the product of the ${{m}_{1}}$ and ${{m}_{2}}$ slopes, which is

${{m}_{1}}{{m}_{2}}= -\dfrac{3}{4}\times \dfrac{4}{3}$

${{m}_{1}}{{m}_{2}}= -1$

W.K.T the lines are perpendicular to each other if the relation ${{m}_{1}}{{m}_{2}}= -1$ is used.

Hence the lines 3x + 4y + 7 = 0 and 28x - 21y + 50 = 0 are perpendicular to each other.

(ii) Is the line x - 3y = 4 perpendicular to the line 3x - y = 7? 

Ans: Given lines are x - 3y = 4 and 3x - y = 7.

Now, $x-3y=4$,

$-3y= -x+4$-------$\times \left( -1 \right)$

$y=\dfrac{x}{3}-\dfrac{4}{3}$

Compare with, $y=mx+c$

$c= -\dfrac{4}{3}$  and  ${{m}_{1}}=\dfrac{1}{3}$

Now for equation $3x-y=7$,

$-y= -3x+7$ -------$\times \left( -1 \right)$

$y=3x-7$

Compare with, $y=mx+c$

$c= -7$ and ${{m}_{2}}=3$

Now consider the product of the ${{m}_{1}}$ and ${{m}_{2}}$ slopes, which is

${{m}_{1}}{{m}_{2}}= \dfrac{1}{3}\times 3$

${{m}_{1}}{{m}_{2}}=1\ne  -1$

Because the lines are perpendicular to each other only if the relation ${{m}_{1}}{{m}_{2}}= -1$ is used.

Hence the lines x - 3y = 4 and 3x - y = 7 are not perpendicular to each other.

(iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1? 

Ans: Given lines are 3x + 2y = 5 and x + 2y = 1.

Now, $3x+2y=5$,

$2y= -3x+5$

$y= -\dfrac{3x}{2}+\dfrac{5}{2}$

Compare with, $y=mx+c$

$c=\dfrac{5}{2}$  and  ${{m}_{1}}= -\dfrac{3}{2}$

Now for equation $x+2y=1$,

$2y= -x+1$ 

$y= -\dfrac{x}{2}+\dfrac{1}{2}$

Compare with, $y=mx+c$

$c=\dfrac{1}{2}$  and  ${{m}_{2}}= -\dfrac{1}{2}$

Now consider the product of the ${{m}_{1}}$ and ${{m}_{2}}$ slopes, which is

${{m}_{1}}{{m}_{2}}=\left( - \dfrac{3}{2} \right)\times \left( -\dfrac{1}{2} \right)$

Here ${{m}_{1}}\ne {{m}_{2}}$, because both the lines are parallel to each other.

(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.

Ans: Let the points are $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 1,4 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( x,2 \right)$.

The slope will be,

$m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$2=\dfrac{2 - 4}{x - 1}$

$2\left( x-1 \right)= -2$

$2x-2= -2$------$\left( \div 2 \right)$

$x-1= -1$

$x=1-1$

$x=0$

4. Find the slope of the line which is parallel to: 

(i) $x+2y+3=0$ 

Ans: The given line is $x+2y+3=0$

$2y= -x-3$

$2y= -\dfrac{x}{2}-\dfrac{3}{2}$

Comparing with, $y=mx+c$

$m= -\dfrac{1}{2}$ and  $c= -\dfrac{3}{2}$

Therefore the slope of line which is parallel to the given line is $-\dfrac{1}{2}$.

(ii) $\dfrac{x}{2}-\dfrac{y}{3}-1=0$

Ans: The given line is $\dfrac{x}{2}-\dfrac{y}{3}-1=0$

$-\dfrac{y}{3}= -\dfrac{x}{2}+1$-------$\times \left( -1 \right)$

$\dfrac{y}{3}=\dfrac{x}{2}-1$

$y=\dfrac{3x}{2}-3$

Comparing with, $y=mx+c$

$m=\dfrac{3}{2}$  and  $c= -3$

Therefore the slope of line which is parallel to the given line is $\dfrac{3}{2}$.

5. Find the slope of the line which is perpendicular to: 

(i) $x-\dfrac{y}{2}+3=0$

Ans: Given, $x-\dfrac{y}{2}+3=0$

$-\dfrac{y}{2}= -x+3$-------$\times \left( -1 \right)$

$\dfrac{y}{2}=x-3$

$y=2x-6$

Comparing with, $y=mx+c$

${{m}_{1}}=2$ and $c= -6$

Slope of the line perpendicular to the given line will be,

${{m}_{1}}{{m}_{2}}= -1$

${{m}_{2}}=\dfrac{-1}{{{m}_{1}}}$

${{m}_{2}}= -\dfrac{1}{2}$

(ii) $\dfrac{x}{3}-2y=4$

Ans: Given, $\dfrac{x}{3}-2y=4$

$-2y= -\dfrac{x}{3}+4$-------$\times \left( -1 \right)$

$2y=\dfrac{x}{3}-4$

$y=\dfrac{x}{6}-\dfrac{4}{2}$

$y=\dfrac{x}{6}-2$

Comparing with, $y=mx+c$

${{m}_{1}}=\dfrac{1}{6}$  and  $c= -2$

Slope of the line perpendicular to the given line will be,

${{m}_{1}}{{m}_{2}}= -1$

${{m}_{2}}=\dfrac{-1}{\dfrac{1}{6}}$

${{m}_{2}}= -6$

6. (i) Lines 2x - by + 5 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b. 

Ans: Given lines 2x - by + 5 = 0 and ax + 3y = 2 

Now, in equation $2x-by+5=0$

$-by= -2x-5$-------$\times \left( -1 \right)$

$by=2x+5$

$y=\dfrac{2x}{b}+\dfrac{5}{b}$

Comparing with, $y=mx+c$

${{m}_{1}}=\dfrac{2}{b}$  and  $c=\dfrac{5}{b}$

And in equation $ax+3y=2$

$3y= -ax+2$

$y= -\dfrac{ax}{3}+\dfrac{2}{3}$

Comparing with, $y=mx+c$

${{m}_{2}}= -\dfrac{a}{3}$  and  $c=\dfrac{2}{3}$

Here, both the lines are parallel to each other. Therefore,

${{m}_{1}}={{m}_{2}}$

$\dfrac{2}{b}= -\dfrac{a}{3}$

$-ab=6$

$ab+6=0$

Hence, the relation connecting a and b is $ab+6=0$

(ii) Lines mx + 3y = -7 and 5x - ny = 3 are perpendicular to each other. Find the relation connecting m and n.

Ans: Given lines mx + 3y = -7 and 5x - ny = 3 

Now, in equation $mx+3y= -7$

$3y= -mx-7$

$y= -\dfrac{mx}{3}-\dfrac{7}{3}$

Comparing with, $y=mx+c$

${{m}_{1}}= -\dfrac{m}{3}$  and  $c= -\dfrac{7}{3}$

And in equation $5x-ny=3$

$-ny= -5x+3$-------$\times \left( -1 \right)$

$ny=5x-3$

$y=\dfrac{5x}{n}-\dfrac{3}{n}$

Comparing with, $y=mx+c$

${{m}_{2}}=\dfrac{5}{n}$  and  $c= -\dfrac{3}{n}$

Here, both the lines are perpendicular to each other. Therefore,

${{m}_{1}}{{m}_{2}}= -1$

$-\dfrac{m}{3}\times \dfrac{5}{n}= -1$

$-\dfrac{5m}{3n}= -1$

$5m=3n$

Hence, the relation connecting a and b is $5m=3n$

7. Find the value of p if the lines, whose equations are 2x - y + 5 = 0 and px + 3y = 4 are perpendicular to each other.

Ans: Given equations are 2x - y + 5 = 0 and px + 3y = 4

Now, in equation $2x-y+5=0$

$-y= -2x-5$-------$\times \left( -1 \right)$

$y=2x+5$

Comparing with, $y=mx+c$

${{m}_{1}}=2$ and $c=5$

And in equation $px+3y=4$

$3y= -px+4$

$y= -\dfrac{px}{3}+\dfrac{4}{3}$

Comparing with, $y=mx+c$

${{m}_{2}}= -\dfrac{p}{3}$  and  $c=\dfrac{4}{3}$

Here, both the lines are perpendicular to each other. Therefore,

${{m}_{1}}{{m}_{2}}= -1$

$2\times \left( -\dfrac{p}{3} \right)= -1$

$-\dfrac{2p}{3}= -1$

$2p=3$

$p=\dfrac{3}{2}$

8. The equation of a line AB is 2x - 2y + 3 = 0. 

(i) Find the slope of the line AB. 

(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.

Ans: Given equation of the line is $2x-2y+3=0$.

(i) To find slope of the line AB

$2x-2y+3=0$

$-2y= -2x-3$------$\times \left( -1 \right)$

$2y=2x+3$

$y=\dfrac{2x}{2}+\dfrac{3}{2}$

$y=x+\dfrac{3}{2}$

Compare with equation of the line, $y=mx+c$

We get slope of the line AB = m = 1

(ii) The angle of line with the positive direction of x-axis will be,

$m=tan \theta $

$1=tan \theta $

$\theta =ta{{n}^{-1}}\left( 1 \right)$

$\theta ={{45}^{o}}$

9. The lines represented by 4x + 3y = 9 and px - 6y + 3 = 0 are parallel. Find the value of p.

Ans: Given lines are 4x + 3y = 9 and px - 6y + 3 = 0

Now, in equation $4x+3y=9$

$3y= -4x+9$

$y= -\dfrac{4x}{3}+\dfrac{9}{4}$

Comparing with, $y=mx+c$

${{m}_{1}}= -\dfrac{4}{3}$  and  $c=\dfrac{9}{4}$

And in equation $px-6y+3=0$

$-6y= -px-3$-------$\times \left( -1 \right)$

$6y=px+3$

$y=\dfrac{px}{6}+\dfrac{3}{6}$

$y=\dfrac{px}{6}+\dfrac{1}{2}$

Comparing with, $y=mx+c$

${{m}_{2}}=\dfrac{p}{6}$  and  $c=\dfrac{1}{2}$

Here, both the lines are parallel to each other. Therefore,

${{m}_{1}}={{m}_{2}}$

$-\dfrac{4}{3}=\dfrac{p}{6}$

$-24=3p$

$p= -\dfrac{24}{3}$

$p= -7$

10. If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p. 

Ans: Given lines are y = 3x + 7 and 2y + px = 3

Now, $y=3x+7$

Comparing with, $y=mx+c$

${{m}_{1}}=3$  and  $c=7$

And in equation $2y+px=3$

$2y= -px+3$

$y= -\dfrac{px}{2}+\dfrac{3}{2}$

Comparing with, $y=mx+c$

${{m}_{2}}= -\dfrac{p}{2}$  and  $c=\dfrac{3}{2}$

Here, both the lines are perpendicular to each other. Therefore,

${{m}_{1}}{{m}_{2}}= -1$

$3\times \left( -\dfrac{p}{2} \right)= -1$

$-\dfrac{3p}{2}= -1$

$3p=2$

$p=\dfrac{2}{3}$

11. The line through A(-2, 3) and B(4, b) is Perpendicular to the line 2x - 4y = 5. Find the value of b.

Ans: The given points are A = (-2, 3) and B = (4, b).

Slope of line passing through the points A and B will be

${{m}_{1}}=\dfrac{b - 3}{4 - \left( -2 \right)}$

${{m}_{1}}=\dfrac{b - 3}{6}$

The given equation of the line is $2x-4y=5$,

$-4y = -2x+5$------$\times \left( -1 \right)$

$4y =2x-5$

$y =\dfrac{2x}{4}-\dfrac{5}{2}$

$y =\dfrac{x}{2}-\dfrac{5}{2}$

Comparing with, $y=mx+c$

${{m}_{2}}=\dfrac{1}{2}$  and  $c= -\dfrac{5}{2}$

Here, both the lines are perpendicular to each other. Therefore,

${{m}_{1}}{{m}_{2}}= -1$

$\left( \dfrac{b - 3}{6} \right)\times \left( \dfrac{1}{2} \right)= -1$

$\dfrac{b - 3}{12}= -1$

$b-3= -12$

$b= -9$

12. Find the equation of the line passing through (-5, 7) and parallel to: 

(i) x-axis 

(ii) y-axis

Ans: 

(i) Slope of the line parallel to the x-axis is 0.

Let the given points be $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -5,7 \right)$.

Equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-7=0\left( x-\left( -5 \right) \right)$

$y-7=0$

$y=7$

(ii) The slope of the line parallel to the x-axis is not defined.

So, the given equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$x-{{x}_{1}}=0$

$x-\left( -5 \right)=0$

$x+5=0$

13. (i) Find the equation of the line passing through (5, -3) and parallel to x - 3y = 4.

(ii) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1).

Ans: 

(i) Given equation of the line is $x-3y=4$

$-3y= -x+4$------$\times \left( -1 \right)$

$3y=x-4$

$y=\dfrac{x}{3}-\dfrac{4}{3}$

Comparing with, $y=mx+c$

${{m}_{2}}=\dfrac{1}{3}$  and  $c= -\dfrac{4}{3}$

The equation of the line passing through point (5, -3) and parallel to the given equation will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-\left( -3 \right)=\dfrac{1}{3}\left( x-5 \right)$

$3\left( y+3 \right)=\left( x-5 \right)$

$3y+9=x-5$

$x-3y-14=0$

(ii) Given equation of the line is, $3x+2y=8$

$2y= -3x+8$

$y= -\dfrac{3}{2}x+\dfrac{8}{2}$

$y= -\dfrac{3}{2}x+4$

Comparing with, $y=mx+c$

${{m}_{2}}= -\dfrac{3}{2}$  and  $c=4$

The equation of the line passing through point (0, 1) and parallel to the given equation will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-1= -\dfrac{3}{2}\left( x-0 \right)$

$2\left( y-1 \right)= -3\left( x-0 \right)$

$2y-2= -3x$

$3x+2y-2=0$

14. Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.

Ans: Given equation of the line is, $4x+5y=6$

$5y= -4x+6$

$y= -\dfrac{4}{5}x+\dfrac{6}{5}$

Comparing with, $y=mx+c$

${{m}_{2}}= -\dfrac{4}{5}$  and  $c=\dfrac{6}{5}$

Slope of line perpendicular to the given line is,

${{m}_{1}}{{m}_{2}}= -1$

${{m}_{1}}\left( -\dfrac{4}{5} \right)= -1$

${{m}_{1}}=\dfrac{5}{4}$

Equation of the line passing through the point (−2, 1) will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-1=\dfrac{5}{4}\left( x-\left( -2 \right) \right)$

$4\left( y-1 \right)=5\left( x+2 \right)$

$4y-4=5x+10$

$5x-4y+14=0$

15. Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).

Ans: Given points are A = (6, -3) and B = (0, 3).

The point that bisects the given points will have the coordinates,

$\left( {{x}_{1}},{{y}_{1}} \right)=\left( \dfrac{{{x}_{1}} + {{x}_{2}}}{2},\dfrac{{{y}_{1}} + {{y}_{2}}}{2} \right)$

$\left( {{x}_{1}},{{y}_{1}} \right)=\left( \dfrac{6 + 0}{2},\dfrac{-3 + 3}{2} \right)$

$\left( {{x}_{1}},{{y}_{1}} \right)=\left( \dfrac{6}{2},\dfrac{0}{2} \right)$

$\left( {{x}_{1}},{{y}_{1}} \right)=\left( 3,0 \right)$

Slope of AB will be,

${{m}_{AB}}=\dfrac{3 - \left( -3 \right)}{0 - 6}$

${{m}_{AB}}=\dfrac{6}{-6}$

${{m}_{AB}}= -1$

Slope of line perpendicular to AB will be,

$m {{m}_{AB}}= -1$

$m=\dfrac{-1}{{{m}_{AB}}}$

$m=\dfrac{-1}{-1}$

$m=1$

Equation of the perpendicular bisector will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-0=1\left( x-3 \right)$

$y=x-3$

16. In the following diagram, write down: 

(i) the coordinates of the points A, B and C. 

(ii) the equation of the line through A and parallel to BC.

Ans: 

(i) From the diagram,

The coordinate of point A = (2, 3), B = (-1, 2) and C = (3, 0)

(ii) Slope of BC is,

${{m}_{BC}}=\dfrac{0 - 2}{3 - \left( -1 \right)}$

${{m}_{BC}}=\dfrac{-2}{4}$

${{m}_{BC}}=\dfrac{-1}{2}$

Slope of line parallel to BC is,

$m={{m}_{BC}}$

$m=\dfrac{-1}{2}$

Equation of the line passing through A and parallel to BC is,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-3= -\dfrac{1}{2}\left( x-2 \right)$

$2\left( y-3 \right)= -1\left( x-2 \right)$

$2y-6= -x+2$

$x+2y=8$

17. B (-5. 6) and D (1, 4) are the vertices of rhombus ABCD. Find the equations of diagonals BD and AC.

Ans: Given vertices of the rhombus are B = (-5. 6) and D = (1, 4)

The diagonals of a rhombus bisect each other at right angles. Therefore, the point where diagonals AC and BD intersect will be,

$E=\left( \dfrac{-5 + 1}{2},\dfrac{6 + 4}{2} \right)$

$E=\left( \dfrac{-4}{2},\dfrac{10}{2} \right)$

$E=\left( -2,5 \right)$

Slope of diagonal BD will be,

${{m}_{BD}}=\dfrac{4 - 6}{1 - \left( -5 \right)}$

${{m}_{BD}}=\dfrac{-2}{6}$

${{m}_{BD}}=\dfrac{-1}{3}$

Slope of diagonal AC will be,

${{m}_{BD}} {{m}_{AC}}= -1$

${{m}_{AC}}\left( \dfrac{-1}{3} \right)= -1$

${{m}_{AC}}=3$

Equation of the diagonal BD is

$y-{{y}_{1}}={{m}_{BD}}\left( x-{{x}_{1}} \right)$

$y-6= -\dfrac{1}{3}\left( x-\left( -5 \right) \right)$

$3\left( y-6 \right)= -1\left( x+5 \right)$

$3y-18= -x-5$

$x+3y=13$

Equation of the diagonal AC that passes through point E is,

$y-{{y}_{1}}={{m}_{AC}}\left( x-{{x}_{1}} \right)$

$y-5=3\left( x-\left( -2 \right) \right)$

$y-5=3x+6$

$y=3x+11$

18. A = (7, -2) and C = (-1, -6) are the vertices of square ABCD. Find the equations of diagonals AC and BD.

Ans: Given vertices of the square are A = (7, -2) and C = (-1, -6).

Square diagonals cut each other in half at a right angle. So, the position where diagonals AC and BD intersect will be,

$E=\left( \dfrac{7 + \left( -1 \right)}{2},\dfrac{-2 + \left( -6 \right)}{2} \right)$

$E=\left( \dfrac{6}{2},\dfrac{-8}{2} \right)$

$E=\left( 3, -4 \right)$

So the slope of diagonal AC will be,

${{m}_{AC}}=\dfrac{-6 - \left( -2 \right)}{-1 - 7}$

${{m}_{AC}}=\dfrac{-4}{-8}$

${{m}_{AC}}=\dfrac{1}{2}$

Slope of diagonal BD will be,

${{m}_{BD}} {{m}_{AC}}= -1$

${{m}_{BD}}\times \dfrac{1}{2}= -1$

${{m}_{BD}}= -2$

Equation of the diagonal AC is,

$y-{{y}_{1}}={{m}_{AC}}\left( x-{{x}_{1}} \right)$

$y-\left( -2 \right)=\dfrac{1}{2}\left( x-7 \right)$

$2\left( y+2 \right)=\left( x-7 \right)$

$2y+4=x-7$

$2y=x-11$

Equation of the diagonal BD that passes through point E is,

$y-{{y}_{1}}={{m}_{BD}}\left( x-{{x}_{1}} \right)$

$y-\left( -4 \right)= -2\left( x-3 \right)$

$y+4= -2x+6$

$2x+y=$2

19. A (1, -5), B (2, 2) and C (-2, 4) are the vertices of triangle ABC. find the equation of 

(i) the median of the triangle through A. 

(ii) the altitude of the triangle through B.

(iii) the line through C and parallel to AB.

Ans: 

(i) Given vertices are A = (1, 5), B = (2, 2) and C = (-2, 4).

Now, let the median through A bisect BC at point D.

Therefore the coordinate of point D will be,

$D=\left( \dfrac{2 + \left( -2 \right)}{2},\dfrac{2 + 4}{2} \right)$

$D=\left( \dfrac{0}{2},\dfrac{6}{2} \right)$

$D=\left( 0,3 \right)$

Slope of AD will be,

${{m}_{AD}}=\dfrac{3 - \left( -5 \right)}{0 - 1}$

${{m}_{AD}}=\dfrac{8}{- 1}$

${{m}_{AD}}= -8$

Equation of median AD is,

$y-{{y}_{1}}={{m}_{AD}}\left( x-{{x}_{1}} \right)$

$y-\left( -5 \right)= -8\left( x-1 \right)$

$y+5= -8x+8$

$8x+y=3$

(ii) Through B, the altitude will be perpendicular to AC.

As a result, the AC slope will be,

${{m}_{AC}}=\dfrac{4 - \left( -5 \right)}{-2 - 1}$

${{m}_{AC}}=\dfrac{9}{-3}$

${{m}_{AC}}= -3$

Slope of the altitude will be,

$m {{m}_{AC}}= -1$

$m=\dfrac{-1}{{{m}_{AC}}}$

$m=\dfrac{-1}{-3}$

$m=\dfrac{1}{3}$

Equation altitude through B is,

$y-{{y}_{1}}={{m}_{AD}}\left( x-{{x}_{1}} \right)$

$y-2=\dfrac{1}{3}\left( x-2 \right)$

$3\left( y-2 \right)=x-2$

$3y-6=x-2$

$3y=x+4$

(iii) Slope of AB will be,

${{m}_{AB}}=\dfrac{2 - \left( -5 \right)}{2 - 1}$

${{m}_{AB}}=7$

Slope of the line parallel to AB will be,

$m={{m}_{AB}}$

$m=7$

The equation for a line that passes parallel to AB through point C is,

$y-{{y}_{1}}={{m}_{AD}}\left( x-{{x}_{1}} \right)$

$y-4=7\left( x-\left( -2 \right) \right)$

$y-4=7x+14$

$y=7x+18$

20. (i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.

(ii) AB meets the x-axis at A and the y-axis at B. Write down the coordinates of A and B. Calculate the area of triangle OAB, where O is the origin. 

Ans: 

(i) Given point is (3, 2) and the line is 2y = 3x + 5.

Hence the equation is,

$2y=3x+5$

$y=\dfrac{3}{2}x+\dfrac{5}{2}$

Compare with $y=mx+c$

So, slope = $m=\dfrac{3}{2}$

Now, the slope of line perpendicular to the given line is,

$m' m= -1$

$m' \left( \dfrac{3}{2} \right)= -1$

$m'= -\dfrac{2}{3}$

Equation of the line passing through the given point is,

$y-{{y}_{1}}=m' \left( x-{{x}_{1}} \right)$

$y-2= -\dfrac{2}{3}\left( x-3 \right)$

$3\left( y-2 \right)= -2\left( x-3 \right)$

$3y-6= -2x+6$

$2x+3y=12$

(ii) For x-axis, Let the coordinate of point A be (x, 0)

Now, substitute y = 0 in the equation of line AB.

$2x+3\times 0=12$

$2x=12$

$x=\dfrac{12}{2}$

$x=6$

Therefore the coordinate of point A = (6, 0).

Now, for the y-axis, Let the coordinate of point A be (0, y).

Substitute x = 0 in the equation of line AB.

$2\times 0+3y=12$

$3y=12$

$y=\dfrac{12}{3}$

$y=4$

Therefore the coordinate of point B = (0, 4).

Hence, OA = 6 and OB = 4.

Therefore, Area of $\Delta $OAB will be,

Area = $\dfrac{1}{2}\times OA\times OB$

Area = $\dfrac{1}{2}\times 6\times 4$

Area = $12$sq units

21. The line 4x - 3y + 12 = 0 meets x-axis at A. Write the coordinates of A. 

Determine the equation of the line through A and perpendicular to 4x - 3y + 12 = 0. 

Ans: Given equation is 4x - 3y + 12 = 0

Let the coordinate of point A be (x, 0)

Now, substitute y = 0 in the given equation.

$4x-3\times 0+12=0$

$4x= -12$

$y=\dfrac{-12}{4}$

$y= -3$

Therefore the coordinate of point A = (−3, 0).

Now, equate the given equation and compare with y = mx + c

$4x-3y+12=0$

$-3y= -4x-12$------$\times \left( -1 \right)$

$3y=4x+12$

$y=\dfrac{4}{3}x+\dfrac{12}{3}$

$y=\dfrac{4}{3}x+4$

Hence slope = $m=\dfrac{4}{3}$

Thus, the slope of line perpendicular to the given line is,

$m' m= -1$

$m' \times \dfrac{4}{3}= -1$

$m'= -\dfrac{3}{4}$

Equation of the line passing through the point A is,

$y-{{y}_{1}}=m' \left( x-{{x}_{1}} \right)$

$y-0= -\dfrac{3}{4}\left( x-\left( -3 \right) \right)$

$4y= -3\left( x+3 \right)$

$4y= -3x-9$

$3x+4y+9=0$

22. The point P is the foot of perpendicular from A (-5, 7) to the line 2x - 3y + 18=0

Determine: 

(i) the equation of the line AP 

(ii) the coordinates of P

Ans: Given point is A = (-5, 7) and the line is 2x - 3y + 18 = 0

Now, equate the given equation and compare with y = mx + c

$2x-3y+18=0$

$-3y= -2x-18$------$\times \left( -1 \right)$

$3y=2x+18$

$y=\dfrac{2}{3}x+\dfrac{18}{3}$

$y=\dfrac{2}{3}x+6$

Hence slope = $m=\dfrac{2}{3}$

Thus, the slope of line perpendicular to the given line is,

$m' m= -1$

$m' \times \dfrac{2}{3}= -1$

$m'= -\dfrac{3}{2}$

Equation of the line passing through the point A is,

$y-{{y}_{1}}=m' \left( x-{{x}_{1}} \right)$

$y-7= -\dfrac{3}{2}\left( x-\left( -5 \right) \right)$

$2\left( y-7 \right)= -3\left( x+5 \right)$

$2y-14= -3x-15$

$3x+2y+1=0$

(ii) The intersection of the given line and line AP is termed as P.

So, the given line is, 2x - 3y + 18 = 0 ------ (1) 

And, 3x + 2y + 1 = 0 ------ (2)

Multiply equation (1) by 2 and equation (2) by 3

4x - 6y + 36 = 0 ------ (3) 

And, 9x + 6y + 3 = 0 ------ (4)

Adding equation (3) and equation (4). We get,

$4x-6y+36+9x+6y+3=0$

$13x+39=0$

$13x= -39$

$x=\dfrac{-39}{13}$

$x= -3$

Substitute x = −3 in equation (2).

$3\left( -3 \right)+2y+1=0$

$-9+2y+1=0$

$2y-8=0$

$2y=8$

$y=\dfrac{8}{2}$

$y=4$

Therefore the coordinate of point P will be (−3, 4).

23. The points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC.

If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the coordinates of P and Q.

Ans: Given points are A = (4, 0), B = (2, 2) and C = (0, 6).

Slope of AB will be,

${{m}_{AB}}=\dfrac{2 - 0}{2 - 4}$

${{m}_{AB}}=\dfrac{2}{- 2}$

${{m}_{AB}}= -1$

Slope of BC will be,

${{m}_{BC}}=\dfrac{6 - 2}{0 - 2}$

${{m}_{BC}}=\dfrac{4}{- 2}$

${{m}_{BC}}= -2$

Equation of AB will be,

$y-{{y}_{1}}={{m}_{AB}} \left( x-{{x}_{1}} \right)$

$y-0= -1\left( x-4 \right)$

$y= -x+4$

$x+y=4$

Equation of BC will be,

$y-{{y}_{1}}={{m}_{BC}} \left( x-{{x}_{1}} \right)$

$y-2= -2\left( x-2 \right)$

$y-2= -2x+4$

$2x+y=6$

Let the coordinate of point P be (0, y).

Now, substitute x = 0 in the equation of AB,

0 + y = 4

y = 4

Therefore the coordinate of point P = (0, 4).

Let the coordinate of point Q is (x, 0).

Substitute y = 0 in the equation of BC,

$2x+0=6$

$x=\dfrac{6}{2}$

$x=3$

Therefore the coordinate of point Q will be (3, 0).

24. Match the equations A, B, C and D with the lines ${{L}_{1}},{{L}_{2}},{{L}_{3}}$ and ${{L}_{4}}$, whose graphs are roughly drawn in the given diagram.

$A\equiv y=2x$

$B\equiv y-2x+2=0$

$C\equiv 3x+2y=6$

$D\equiv y=2$

Ans: For A, $y=2x$

Now, substitute $x=0$ in the above equation

$y=2\times 0$

$y=0$

Hence, the line passes through the origin.

Thus, A → ${{L}_{3}}$

For B, $y-2x+2=0$

Now, substitute $x=0$ in the above equation

$y-2\times 0+2=0$

$y= -2$

Substitute, $y=0$

$0-2x+2=0$

$-2x= -2$

$x=\dfrac{-2}{-2}$

$x=1$

Here, x-intercept is positive and y-intercept is negative.

Thus, B → ${{L}_{4}}$

For C, $3x+2y=6$

Now, substitute $x=0$ in the above equation

$3\times 0+2y=6$

$2y=6$

$y=\dfrac{6}{2}$

$y=3$

Substitute, y = 0

$3x+2\times 0=6$

$3x=6$

$y=\dfrac{6}{3}$

$y=2$

Here, x-intercept and y-intercept both are positive.

Thus, C → ${{L}_{2}}$

For D, $y=2$

This is the equation of line parallel to the x-axis.

Thus, D → ${{L}_{1}}$

25. Find the value of ‘a’ for which the points A(a, 3), B(2, 1) and C(5, a) are collinear. Hence, find the equation of the line.

Ans: The given points are A = (a, 3), B = (2, 1) and C = (5, a).

Slope of the AB will be,

${{m}_{AB}}=\dfrac{1 - 3}{2 - a}$

${{m}_{AB}}=\dfrac{-2}{2 - a}$

Slope of the BC will be,

${{m}_{BC}}=\dfrac{a - 1}{5 - 2}$

${{m}_{BC}}=\dfrac{a - 1}{3}$

Since, the points are collinear.

${{m}_{AB}}={{m}_{BC}}$

$\dfrac{-2}{2 - a}=\dfrac{a - 1}{3}$

On equating the above equation,

$3\left( -2 \right)=\left( a-1 \right)\left( 2-a \right)$

$-6=2a-{{a}^{2}}-2+a$

${{a}^{2}}-3a-4=0$

On factorising we get,

${{a}^{2}}-4a+a-4=0$

$a\left( a-4 \right)+1\left( a-4 \right)=0$

$\left( a-4 \right)\left( a+1 \right)=0$

Therefore, $a=4, -1$

Now, to find the equation of the line,

Taking $a=4$

Slope will be,

$m=\dfrac{a - 1}{3}$

$m=\dfrac{4 - 1}{3}$

$m=\dfrac{3}{3}$

$m=1$

Thus, the equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-1=1\left( x-2 \right)$

$y=x-2+1$

$y=x-1$

Hence, $x-y=1$

Now, take $a= -1$

Slope will be,

$m=\dfrac{a - 1}{3}$

$m=\dfrac{-1 - 1}{3}$

$m=\dfrac{-2}{3}$

Thus, the equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-1=\dfrac{-2}{3}\left( x-2 \right)$

$3\left( y-1 \right)= -2\left( x-2 \right)$

$3y-3= -2x+4$

Hence, $2x+3y=7$..

EXERCISE 14 (E)

1. Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3 : 5. Find its coordinates of point P. 

Also, find the equation of the line through P and parallel to 3x + 5y = 7.

Ans: Given points are A = (8, 0) and B = (16, -8).

By using the section formula, the coordinate of point P will be,

$P=\left( \dfrac{m{{x}_{2}} + n{{x}_{1}}}{m + n},\dfrac{n{{y}_{2}} + n{{y}_{1}}}{m + n} \right)$

$P=\left( \dfrac{3 \times  16 + 5 \times  8}{3 + 5},\dfrac{3 \times  \left( -8 \right) + 5 \times  0}{3 + 5} \right)$

$P=\left( \dfrac{48 + 40}{8},\dfrac{-24}{8} \right)$

$P=\left( \dfrac{88}{8},-3 \right)$

$P=\left( 11,-3 \right)$

Now, equate the given equation and compare with $y=mx+c$

$3x+5y=7$

$5y= -3x+7$

$y= -\dfrac{3}{5}x+\dfrac{7}{5}$

Hence slope = $m= -\dfrac{3}{5}$

Thus, the equation of the line through point P is,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-\left( -3 \right)= -\dfrac{3}{5}\left( x-11 \right)$

$5\left( y+3 \right)= -3\left( x-11 \right)$

$5y+15= -3x+33$

$3x+5y=18$

2. The line segment joining the points A (3, -4) and B (-2, 1) is divided in the ratio 1 : 3 at point P in it. Find the coordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x - 3y = 4.

Ans: Given points are A = (3, -4) and B = (-2, 1).

By using the section formula, the coordinate of point P will be,

$P=\left( \dfrac{m{{x}_{2}} + n{{x}_{1}}}{m + n},\dfrac{n{{y}_{2}} + n{{y}_{1}}}{m + n} \right)$

$P=\left( \dfrac{1 \times  \left( -2 \right) + 3 \times  3}{1 + 3},\dfrac{1 \times  1 + 3 \times  \left( -4 \right)}{1 + 3} \right)$

$P=\left( \dfrac{-2 + 9}{4},\dfrac{1 - 12}{4} \right)$

$P=\left( \dfrac{7}{4},\dfrac{-11}{4} \right)$

Now, equate the given equation and compare with $y=mx+c$

$5x-3y=4$

$-3y= -5x+4$

$y=\dfrac{-5}{-3}x+\left( \dfrac{4}{-3} \right)$

$y=\dfrac{5}{3}x-\dfrac{4}{3}$

Hence slope = $m=\dfrac{5}{3}$

The slope of line perpendicular to the given line will be

$m'm= -1$

$m'\times \dfrac{5}{3}= -1$

$m'= -\dfrac{3}{5}$

Thus, the equation of the line through point P is,

$y-{{y}_{1}}=m'\left( x-{{x}_{1}} \right)$

$y-\left( -\dfrac{11}{4} \right)= -\dfrac{3}{5}\left( x-\dfrac{7}{4} \right)$

$5\left( \dfrac{4y + 11}{4} \right)= -3\left( \dfrac{4x - 7}{4} \right)$

$5\left( 4y+11 \right)= -3\left( 4x-7 \right)$

$20y+55= -12x+21$

$12x+20y+34=0$------$\left( \div 2 \right)$

$6x+10y+17=0$

3. A line 5x + 3y + 15 = 0 meets the y-axis at point P. Find the coordinates of point P. Find the equation of a line through P and perpendicular to x - 3y + 4 = 0.

Ans: Point P lies on the y-axis, so substitute x = 0 in equation 5x + 3y + 15 = 0

$5\times 0+3y+15=0$

$3y+15=0$

$3y= -15$

$y= -\dfrac{15}{3}$

$y= -5$

Hence, the coordinate of point P is (0, -5).

Now, The slope of the line x - 3y + 4 = 0 will be,

$x-3y+4=0$

$-3y= -x-4$------$\times \left( -1 \right)$

$3y=x+4$

$y=\dfrac{x}{3}+\dfrac{4}{3}$

Comparing with, y = mx + c,

Slope = $m=\dfrac{1}{3}$

Slope of line perpendicular to the given line will be,

$m'm= -1$

$m'\times \dfrac{1}{3}= -1$

$m'= -3$

Thus, the equation of the line is,

$y-{{y}_{1}}=m'\left( x-{{x}_{1}} \right)$

$y-\left( -5 \right)= -3\left( x-0 \right)$

$y+5= -3x$

$3x+y+5=0$

4. Find the value of ‘k’ for which the lines kx - 5y + 4 = 0 and 5x - 2y + 5 = 0 are perpendicular to each other.

Ans: Slope of the line kx - 5y + 4 = 0 on compare with y = mx + c will be,

$kx-5y +4=0$

$-5y= -kx-4$------$\times \left( -1 \right)$

$5y=kx+4$

$y=\dfrac{k}{5}x+\dfrac{4}{5}$

Hence slope = ${{m}_{1}}=\dfrac{k}{5}$

Now, slope of the line 5x - 2y + 5 = 0 on comparing with y = mx + c will be,

$5x-2y+5=0$

$-2y= -5x-5$------$\times \left( -1 \right)$

$2y=5x+5$

$y=\dfrac{5}{2}x+\dfrac{5}{2}$

Hence slope = ${{m}_{2}}=\dfrac{5}{2}$

Since, both the lines are perpendicular to each other.

Thus ${{m}_{1}}{{m}_{2}}= -1$

$\dfrac{k}{5}\times \dfrac{5}{2}= -1$

$\dfrac{k}{2}= -1$

$k= -2$

5. A straight line passes through the points P(-1, 4) and Q(5, -2). It intersects the coordinate axes at points A and B. M is the midpoint of the segment AB. Find: 

(i) The equation of the line.

(ii) The coordinates of A and B.

(iii) The coordinates of M.

Ans: 

(i) The given points are P = (-1, 4) and Q = (5, -2).

Slope of PQ will be,

$m=\dfrac{{{y}_{2}} - {{y}_{1}}}{{{x}_{2}} - {{x}_{1}}}$

$m=\dfrac{-2 - 4}{5 - \left( -1 \right)}$

$m=\dfrac{-6}{6}$

$m= -1$

Equation of line passing through PQ will be,

$y-{{y}_{1}}=m\left( \text{x}-{{x}_{1}} \right)$

$y-4= -1\left( \text{x}-\left( -1 \right) \right)$

$y-4= -x-1$

$x+y=3$

(ii) Let the coordinates of point A and B be (x, 0) and (0, y) respectively.

Substitute y = 0 in equation of line,

$x+0=3$

$x=3$

Substitute x = 0 in equation of line,

$0+y=3$

$y=3$

Hence, the coordinate of point A = (3, 0) and B = (0, 3).

(iii) Point M is the midpoint of AB.

Thus, the coordinate of point M will be,

$M=\left( \dfrac{{{y}_{2}} + {{y}_{1}}}{2},\dfrac{{{x}_{2}} + {{x}_{1}}}{2} \right)$

$M=\left( \dfrac{3 + 0}{2},\dfrac{0 + 3}{2} \right)$

$M=\left( \dfrac{3}{2},\dfrac{3}{2} \right)$

6. (1, 5) and (-3, -1) are the coordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD. 

Ans: Given vertices of the rhombus are A = (1, 5) and C = (-3, -1)

Diagonals in a rhombus bisect each other at a right angle. As a result, the position where diagonals AC and BD intersect will be,

$E=\left( \dfrac{1 + \left( -3 \right)}{2},\dfrac{5 + \left( -1 \right)}{2} \right)$

$E=\left( \dfrac{-2}{2},\dfrac{4}{2} \right)$

$E=\left( -1,2 \right)$

Slope of diagonal AC will be,

${{m}_{AC}}=\dfrac{-1 - 5}{-3 - 1}$

${{m}_{AC}}=\dfrac{-6}{-4}$

${{m}_{AC}}=\dfrac{3}{2}$

Slope of diagonal BD will be,

${{m}_{BD}} {{m}_{AC}}= -1$

${{m}_{BD}}\times \dfrac{3}{2}= -1$

${{m}_{BD}}= -\dfrac{2}{3}$

Equation of the diagonal AC is,

$y-{{y}_{1}}={{m}_{AC}}\left( x-{{x}_{1}} \right)$

$y-5=\dfrac{3}{2}\left( x-1 \right)$

$2\left( y-5 \right)=3\left( x-1 \right)$

$2y-10=3x-3$

$3x-2y+7=0$

Equation of the diagonal BD that passes through point E is,

$y-{{y}_{1}}={{m}_{BD}}\left( x-{{x}_{1}} \right)$

$y-2= -\dfrac{2}{3}\left( x-\left( -1 \right) \right)$

$3\left( y-2 \right)= -2\left( x+1 \right)$

$3y-6= -2x-2$

$2x+3y-4=0$

7. Show that A (3, 2), B (6, -2) and C (2. -5), can be the vertices of a square. 

(i) Find the coordinates of its fourth vertex D, if ABCD is a square. 

(ii) Without using the coordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.

Ans: The given vertices are A = (3, 2), B = (6, -2) and C = (2. -5)

Slope of AB is,

${{m}_{AB}}=\dfrac{-2 - 2}{6 - 3}$

${{m}_{AB}}=\dfrac{-4}{3}$

Slope of BC is,

${{m}_{BC}}=\dfrac{-5 - \left( -2 \right)}{2 - 6}$

${{m}_{BC}}=\dfrac{-3}{-4}$

${{m}_{BC}}=\dfrac{3}{4}$

Hence, Slope of AB $\times $ Slope of BC = $-1$

Thus, AB $\bot $ BC,

Therefore the A, B and C can be the vertices of the square.

(i) As we know that the diagonals of a square bisect each other at right angles.

Therefore the midpoint on diagonal AC will be,

$E=\left( \dfrac{3 + 2}{2},\dfrac{2 + \left( -5 \right)}{2} \right)$

$E=\left( \dfrac{5}{2},\dfrac{-3}{2} \right)$

Also point E bisects the diagonal BD,

$E=\left( \dfrac{6 + x}{2},\dfrac{-2 + y}{2} \right)$

$\left( \dfrac{5}{2},\dfrac{-3}{2} \right)=\left( \dfrac{6 + x}{2},\dfrac{-2 + y}{2} \right)$

Comparing both the sides we get,

$\dfrac{5}{2}=\dfrac{6 + x}{2}$

$2\times 5=2\left( 6+x \right)$

$10=12+2x$

$2x=10-12$

$2x= -2$

$x=\dfrac{-2}{2}$

$x= -1$

And,

$\dfrac{-3}{2}=\dfrac{-2 + y}{2}$

$2\times \left( -3 \right)=2\left( -2+y \right)$

$-6= -4+2y$

$2y= -6+4$

$2x= -2$

$x=\dfrac{-2}{2}$

$x= -1$

Hence, coordinate of point D = (-1, -1).

(ii) In the square ABCD, AD is parallel to BC.

Slope of BC is known i.e., $\dfrac{3}{4}$

As AD is parallel to BC, Slope of BC = Slope of AD = $\dfrac{3}{4}$

Equation of side AD will be,

$\left( y-{{y}_{1}} \right)={{m}_{AD}}\left( x-{{x}_{1}} \right)$

$\left( y-2 \right)=\dfrac{3}{4}\left( x-3 \right)$

$4\left( y-2 \right)=3\left( x-3 \right)$

$4y-8=3x-9$

$4y=3x-1$

Slope of AC will be,

${{m}_{AC}}=\dfrac{-5 - 2}{2 - 3}$

${{m}_{AC}}=\dfrac{-7}{-1}$

${{m}_{AC}}=7$

Slope of BD will be,

${{m}_{BD}} {{m}_{AC}}= -1$

${{m}_{BD}} = -\dfrac{1}{7}$

Equation of diagonal BD will be,

$y-{{y}_{1}}={{m}_{BD}}\left( x-{{x}_{1}} \right)$

$y-\left( -2 \right)= -\dfrac{1}{7}\left( x-6 \right)$

$7\left( y+2 \right)= -1\left( x-6 \right)$

$7y+14= -x+6$

$x+7y+8=0$

8. A line through origin meets the line x = 3y + 2 at right angles at point X. Find the coordinates of X. 

Ans: The slope of the line x = 3y + 2 will be,

$x=3y+2$

$3y=x-2$

$y=\dfrac{1}{3}x-\dfrac{2}{3}$

Compare with y = mx + c,

Slope = $m=\dfrac{1}{3}$

Since, both the lines are perpendicular to each other,

$m'm= -1$

$m'\times \dfrac{1}{3}= -1$

$m'= -3$

Equation of the line passing through the origin will be,

$y=m'x$

$y= -3x$

Substitute y = -3x in the given equation of the line,

$x=3y+2$

$x=3\left( -3x \right)+2$

$x= -9x+2$

$10x=2$

$x=\dfrac{2}{10}$

$x=\dfrac{1}{5}$

And the value of y will be,

$y= -3x$

$y= -3\left( \dfrac{1}{5} \right)$

$y=\dfrac{-3}{5}$

Therefore the coordinate of point X will be $\left( \dfrac{1}{5},-\dfrac{3}{5} \right)$

9. A straight line passes through the point (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.

Ans: Let (a, 0) and (b, 0) be the coordinates of the line's intersection on the x and y-axes, respectively.

Point (3, 2) is the midpoint.

$\left( 3,2 \right)=\left( \dfrac{a + 0}{2},\dfrac{0 + b}{2} \right)$

$\left( 3,2 \right)=\left( \dfrac{a}{2},\dfrac{b}{2} \right)$

Comparing both the sides, we get

$\dfrac{a}{2}=3$

$a=6$

And,

$\dfrac{b}{2}=2$

$b=4$

Therefore the equation of the line will be,

$\dfrac{x}{a}+\dfrac{y}{b}=1$

$\dfrac{x}{6}+\dfrac{y}{4}=1$

$\dfrac{2x + 3y}{12}=1$

$2x+3y=12$

10. Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x - 8y = -23; and perpendicular to the line 4x - 2y = 1.

Ans: Given equations are,

7x + 6y = 71 ------ (1)

And, 5x - 8y = -23 ------ (2)

Multiply equation (1) by 8 and equation (2) by 6.

56x + 48y = 568 ------ (3)

And, 30x - 48y = -138 ------ (4)

Adding equation (3) and equation (4) we get,

$56x+48y+30x-48y=568-138$

$86x=430$

$x=\dfrac{430}{86}$

$x=5$

Substitute the value of ‘x’ in equation (1),

$7x+6y=71$

$7\left( 5 \right)+6y=71$

$35+6y=71$

$6y=71-35$

$6y=36$

$y=\dfrac{36}{6}$

$y=6$

Thus, the coordinate of the intersection is (5, 6).

The slope of the line $4x-2y=1$ will be,

$2y=4x-1$

$y=\dfrac{4x}{2}-\dfrac{1}{2}$

$y=2x-\dfrac{1}{2}$

Slope of the line = m = 2,

Since, the line is perpendicular to the given line.

Therefore slope will be,

$m'm= -1$

$m'\left( 2 \right)= -1$

$m'= -\dfrac{1}{2}$

The equation of the required line will be,

$y-{{y}_{1}}=m'\left( x-{{x}_{1}} \right)$

$y-6= -\dfrac{1}{2}\left( x-5 \right)$

$2\left( y-6 \right)= -\left( x-5 \right)$

$2y-12= -x+5$

$2y+x=17$

11. Find the equation of the line which is perpendicular to the line $\dfrac{x}{a}-\dfrac{y}{b}=1$ at the point where this line meets the y-axis.

Ans: Given equation $\dfrac{x}{a}-\dfrac{y}{b}=1$

On taking LCM we get,

$\dfrac{bx - ay}{ab}=1$

$bx-ay=ab$

$ay=bx-ab$

$y=\dfrac{b}{a}x-b$

Comparing with, y = mx + c

Slope = m = $\dfrac{b}{a}$

Slope of the line perpendicular to the given line is,

$m'm= -1$

$m'\times \dfrac{b}{a}= -1$

$m'= -\dfrac{a}{b}$

Substitute x = 0 in the above equation for the y-axis intersection coordinate,

$\dfrac{0}{a}-\dfrac{y}{b}=1$

$-\dfrac{y}{b}=1$

$y= -b$

Thus the coordinate is (0, -b).

Equation of the line is,

$y-{{y}_{1}}=m'\left( x-{{x}_{1}} \right)$

$y-\left( -b \right)= -\dfrac{a}{b}\left( x-0 \right)$

$b\left( y+b \right)= -ax$

$by+{{b}^{2}}= -ax$

$ax+by+{{b}^{2}}=0$

12. O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find: 

(i) the equation of median of triangle OAB through vertex O. 

(ii) the equation of altitude of triangle OAB through vertex B. 

Ans: (i) The given vertices of the triangle are O = (0, 0), A = (3, 5) and B = (-5, -3)

Let point C be the mid-point on AB.

Therefore the coordinate of point C will be,

$C=\left( \dfrac{3 + \left( -5 \right)}{2},\dfrac{5 + \left( -3 \right)}{2} \right)$

$C=\left( \dfrac{-2}{2},\dfrac{2}{2} \right)$

$C=\left( -1,1 \right)$

Slope of the median OC will be,

${{m}_{OC}}=\dfrac{1 - 0}{-1 - 0}$

${{m}_{OC}}= -1$

Equation of the median OC will be,

$y=mx$

$y= -x$

$x+y=0$

(ii) Through vertex B, the altitude will be perpendicular to OA. Therefore, the OA slope will be.

${{m}_{OA}}=\dfrac{5 - 0}{3 - 0}$

${{m}_{OA}}=\dfrac{5}{3}$

Slope of the altitude will be,

$m {{m}_{OA}}= -1$

$m\times \dfrac{5}{3}= -1$

$m= -\dfrac{3}{5}$

Equation of the altitude will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-\left( -3 \right)= -\dfrac{3}{5}\left( x-\left( -5 \right) \right)$

$5\left( y+3 \right)= -3\left( x+5 \right)$

$5y+15= -3x-15$

$3x+5y+30=0$

13. Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1. 

Does line 3x = y + 1 bisect the line segments joining the two given points?

Ans: Let the given points are A = (-2, 3) and B = (4, 1)

Slope of AB will be,

$m=\dfrac{1 - 3}{4 - \left( -2 \right)}$

$m=\dfrac{-2}{6}$

$m=\dfrac{-1}{3}$

Slope of the given line will be,

$3x=y+1$

$y=3x-1$

Comparing with, y = mx + c

Slope = $m'=3$

Therefore, $mm'= -1$

Hence, the line through the points A and B is perpendicular to the given line.

Equation of the line AB will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-3= -\dfrac{1}{3}\left( x-\left( -2 \right) \right)$

$3\left( y-3 \right)= -\left( x+2 \right)$

$3y-9= -x-2$

$x+3y=7$------ (1)

And, $3x=y+1$ ------ (2)

Multiply equation (2) with 3

$9x-3y=3$ ------ (3)

Adding equation (1) and (3), we get

$x+3y+9x-3y=7+3$

$10x=10$

$x=1$

Substitute $x$ value in equation (1), we get

$1+3y=7$

$3y=7-1$

$3y=6$

$y=2$

So, the coordinate of the intersection will be (1, 2).

Coordinates of the midpoint of AB are,

$P=\left( \dfrac{-2 + 4}{2},\dfrac{3 + 1}{2} \right)$

$P=\left( \dfrac{2}{2},\dfrac{4}{2} \right)$

$P=\left( 1,2 \right)$

Hence, the given line bisects the line segment AB.

14. Given a straight line x cos 30° + y sin 30° = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3). 

Ans: Given line is x cos 30° + y sin 30° = 2.

W.K.T cos 30° = $\dfrac{\sqrt{3}}{2}$ and y sin 30° = $\dfrac{1}{2}$

Substitute the values in the given equation, we get

$x cos 30{}^\circ +y sin 30{}^\circ =2$

$x \dfrac{\sqrt{3}}{2}+y \dfrac{1}{2}=2$

On taking LCM, we get

$\dfrac{\sqrt{3}x + y}{2}=2$

$\sqrt{3}x+y=4$

$y= -\sqrt{3}x+4$

Comparing with y = mx + c we get,

Slope = $m= -\sqrt{3}$

The equation of the line will be

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-3= -\sqrt{3}\left( x-4 \right)$

$y-3= -\sqrt{3}x+4\sqrt{3}$

$\sqrt{3}x+y=4\sqrt{3}+3$

15. Find the value of k such that the line (k - 2) x + (k + 3) y - 5 =0 is: 

(i) Perpendicular to the line 2x - y + 7 = 0 

(ii) Parallel to it.

Ans: (i) Given line is $\left( k-2 \right)x+\left( k+3 \right)y-5=0$

$\left( k+3 \right)y= -\left( k-2 \right)x+5$

$y= -\dfrac{k - 2}{k + 3} x+\dfrac{5}{k + 3}$

Compare the above equation with y = mx + c

Slope = ${{m}_{1}}= -\dfrac{k - 2}{k + 3}$

And, $2x-y+7=0$

$y=2x+7$

On comparing with y = mx + c,

Slope = ${{m}_{2}}=2$

Here, both the lines are perpendicular to each other,

${{m}_{1}}{{m}_{2}}= -1$

$-\dfrac{k - 2}{k + 3}\times \left( 2 \right)= -1$

$2k-4=k+3$

$2k-k=4+3$

$k=7$

(ii) When both the lines are parallel to each other,

${{m}_{1}}={{m}_{2}}$

$-\dfrac{k - 2}{k + 3}=2$

$-k+2=2k+6$

$-2k-k=6-2$

$-3k=4$

$k= -\dfrac{4}{3}$

16. The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find: 

(i) the equation of line through A and perpendicular to BC. 

(ii) the coordinates of the point P, where the perpendicular through A, as obtained in (i), meets BC. 

Ans: The given vertices of the triangle are A = (0, 5), B = (-1, -2) and C = (11, 7)

Slope of BC will be,

${{m}_{BC}}=\dfrac{7 - \left( -2 \right)}{11 - \left( -1 \right)}$

${{m}_{BC}}=\dfrac{9}{12}$

${{m}_{BC}}=\dfrac{3}{4}$

Equation of BC will be,

$y-{{y}_{1}}={{m}_{BC}}\left( x-{{x}_{1}} \right)$

$y-\left( -2 \right)=\dfrac{3}{4}\left( x-\left( -1 \right) \right)$

$4\left( y+2 \right)=3\left( x+1 \right)$

$4y+8=3x+3$

$3x-4y=5$ ------ (1)

Slope of line perpendicular to BC will be,

$m{{m}_{BC}}= -1$

$m\times \dfrac{3}{4}= -1$

$m= -\dfrac{4}{3}$

Equation of line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-5= -\dfrac{4}{3}\left( x-0 \right)$

$3\left( y-5 \right)= -4x$

$4x+3y-15=0$ ------ (2)

(ii) Multiply equation (1) with 3 and equation (2) with 4, we get

$9x-12y=15$ ------ (3)

And, $16x+12y=60$ ------ (4)

Adding equation (3) and equation (4), we get

$9x-12y+16x+12y=15+60$

$25x=75$

$x=\dfrac{75}{25}$

$x=3$

Substitute ‘x’ value in equation (1), we get 

$3x-4y=5$

$3\times 3-4y=5$

$9-4y=5$

$-4y=5-9$

$-4y= -4$

$y=1$

Thus, the coordinate of the intersection point P is (3, 1).

17. From the given figure, find: 

(i) the coordinates of A, B and C. 

(ii) the equation of the line through A and parallel to BC.

Ans: The coordinates of points A, B and C are: 

A = (2, 3), B = (-1, 2), and C = (3, 0).

(i) Slope of the line parallel to BC = Slope of BC =

${{m}_{BC}}=\dfrac{0 - 2}{3 - \left( -1 \right)}$

${{m}_{BC}}=\dfrac{- 2}{4}$

${{m}_{BC}}= -\dfrac{1}{2}$

(ii) Equation of the line will be,

$y-{{y}_{1}}={{m}_{BC}}\left( x-{{x}_{1}} \right)$

$y-3= -\dfrac{1}{2}\left( x-2 \right)$

$2\left( y-3 \right)= -x+2$

$2y-6= -x+2$

$x+2y=8$

18. P(3, 4), Q(7, -2) and R(-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R.

Ans:  Given vertices of the triangle are P = (3, 4), Q = (7, -2) and R = (-2, -1)

The median through R will bisect PQ. Therefore the mid-point on PQ will be,

$A=\left( \dfrac{3 + 7}{2},\dfrac{4 + \left( -2 \right)}{2} \right)$

$A=\left( \dfrac{10}{2},\dfrac{2}{2} \right)$

$A=\left( 5,1 \right)$

Slope of median RA will be,

$m=\dfrac{1 - \left( -1 \right)}{5 - \left( -2 \right)}$

$m=\dfrac{2}{7}$

Equation of the median RA will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-\left( -1 \right)=\dfrac{2}{7}\left( x-\left( -2 \right) \right)$

$7\left( y+1 \right)=2\left( x+2 \right)$

$7y+7=2x+4$

$7y=2x-3$

19. A(8, -6), B(-4, 2) and C(0, -10) are vertices of a triangle ABC. If P is the midpoint of AB and Q is the midpoint of AC, use coordinate geometry to show that PQ is parallel to BC. Give a special name to quadrilateral PBCQ. 

Ans: Given vertices of the triangle ABC are A = (8, -6), B = (-4, 2) and C = (0, -10)

Mid-point P is,

$P=\left( \dfrac{8 + \left( -4 \right)}{2},\dfrac{-6 + 2}{2} \right)$

$P=\left( \dfrac{4}{2},\dfrac{-4}{2} \right)$

$P=\left( 2,-2 \right)$

Mid-point Q is,

$Q=\left( \dfrac{8 + 0}{2},\dfrac{-6 + \left( -10 \right)}{2} \right)$

$Q=\left( \dfrac{8}{2},\dfrac{-16}{2} \right)$

$Q=\left( 4,-8 \right)$

Slope of PQ is,

${{m}_{PQ}}=\dfrac{-8 - \left( -2 \right)}{4 - 2}$

${{m}_{PQ}}=\dfrac{-6}{2}$

${{m}_{PQ}}= -3$

Slope of BC is,

${{m}_{BC}}=\dfrac{-10 - 2}{0 - \left( -4 \right)}$

${{m}_{BC}}=\dfrac{-12}{4}$

${{m}_{BC}}= -3$

Here, ${{m}_{BC}}={{m}_{PQ}}$

Therefore, PQ is parallel to BC. Hence PBCQ is a trapezium.

20. A line AB meets the x-axis at point A and y-axis at point B. The point P(-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find: 

(i) the coordinates of A and B. 

(ii) equation of line through P and perpendicular to AB.

Ans: (i) Let the coordinates of point A and B are (a, 0) and (0, b) respectively.

Given P = (-4, -2) and AP : PB = 1 : 2.

Using section formula we have,

$\left( -4,-2 \right)=\left( \dfrac{1 \times  0 + 2 \times  a}{1 + 2},\dfrac{1 \times  b + 2 \times  0}{1 + 2} \right)$

$\left( -4,-2 \right)=\left( \dfrac{2a}{3},\dfrac{b}{3} \right)$

Comparing both the sides,

$-4=\dfrac{2a}{3}$

$2a= -12$

$a= -6$

And, $-2=\dfrac{b}{3}$

$b= -6$

Thus, the coordinate of point A is (-6, 0) and point B is (0, -6)

(ii) Slope of AB will be,

${{m}_{AB}}=\dfrac{-6 - 0}{0 - \left( -6 \right)}$

${{m}_{AB}}= -1$

Slope of line perpendicular to the AB will be,

$m{{m}_{AB}}= -1$

$m\left( -1 \right)= -1$

$m=1$

Equation of line through P and perpendicular to AB will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-\left( -2 \right)=1\left( x-\left( -4 \right) \right)$

$y+2=x+4$

$y=x+2$

21. A line intersects the x-axis at point (-2, 0) and cuts off an intercept of 3 units from the positive side of the y-axis. Find the equation of the line.

Ans: The intercept on x-axis at point A = (-2, 0) and y-intercept is 3 which passes through y-axis at B = (0, 3).

Slope of the line AB = $m=\dfrac{3 - 0}{0 + 2}=\dfrac{3}{2}$

Therefore the required equation of the line AB will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-0=\dfrac{3}{2}\left( x-\left( -2 \right) \right)$

$2y=3\left( x+2 \right)$

$2y=3x+6$

22. Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units.

Ans: Let the required line passing through be A = (2, 3)

And the coordinate of the intercept on the x-axis be, B = (4, 0)

The slope of AB will be,

$m=\dfrac{0 - 3}{4 - 2}$

$m=\dfrac{-3}{2}$

Required equation of the line AB will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-0= -\dfrac{3}{2}\left( x-4 \right)$

$2y= -3\left( x-4 \right)$

$2y= -3x+12$

$3x+2y=12$

23. The given figure (not drawn to scale) shows two straight lines AB and CD. If equation of the line AB is: 

$y=x+1$ and equation of line CD is : 

$y=\sqrt{3}x-1$. Write down the inclination of lines AB and CD: also, find the angle $\theta $ between AB and CD. 

Ans: The equation of AB is, $y=x+1$

Comparing with, $y=mx+c$, we get

Slope of AB = m = 1

Inclination of AB $={{45}^{o}}$ (because $tan {{45}^{o}}=1$)

Therefore ${{\theta }_{1}}={{45}^{o}}$

Now, the equation of CD is, $y=\sqrt{3}x-1$

Comparing with, $y=mx+c$, we get

Slope of CD = m = $\sqrt{3}$

Inclination of CD $={{60}^{o}}$ (because $tan {{60}^{o}}=\sqrt{3}$)

Therefore ${{\theta }_{2}}={{60}^{o}}$

Thus, the value of $\theta $ will be,

$\theta ={{\theta }_{2}}-{{\theta }_{1}}$

$\theta ={{60}^{o}}-{{45}^{o}}$

$\theta ={{15}^{o}}$

24. Write down the equation of the line whose gradient is $\dfrac{3}{2}$ and which passes through P, where P divides the line segment joining A(-2, 6) and B(3, -4) in the ratio 2 : 3. 

Ans: The given points are A = (-2, 6) and B = (3, -4) divided by point P in a ratio of 2:3.

The coordinate of point P will be,

$P=\left( \dfrac{2 \times  3 + 3 \times  \left( -2 \right)}{2 + 3},\dfrac{2 \times  \left( -4 \right) + 3 \times  6}{2 + 3} \right)$

$P=\left( \dfrac{6 - 6}{5},\dfrac{-8 + 18}{5} \right)$

$P=\left( 0,2 \right)$

The equation of the line through P will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-2=\dfrac{3}{2}\left( x-0 \right)$

$2\left( y-2 \right)=3x$

$2y-4=3x$

$2y=3x+4$

25. The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the coordinates of that point.

Ans: Let the points are A = (6, 4) and B = (7, -5)

Slope of AB will be,

$m=\dfrac{-5 - 4}{7 - 6}$

$m= -9$

Equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-4= -9\left( x-6 \right)$

$y-4= -9x+54$

$9x+y=58$

Since, the ordinate is -23.

Substitute y = -23 in above equation,

$9x+y=58$

$9x+\left( -23 \right)=58$

$9x=58+23$

$9x=81$

$x=9$

Therefore the coordinate of the point will be (9, -23)

26. Point A and B have coordinates (7, -3) and (1, 9) respectively. Find: 

(i) the slope of AB. 

(ii) the equation of the perpendicular bisector of the line segment AB. 

(iii) the value of ‘p’ if (-2, p) lies on it. 

Ans: (i) The given points are A = (7, -3) and B = (1, 9)

The slope of AB will be,

${{m}_{AB}}=\dfrac{9 - \left( -3 \right)}{1 - 7}$

${{m}_{AB}}=\dfrac{12}{-6}$

${{m}_{AB}}= -2$

(ii) Midpoint of AB will be,

$P=\left( \dfrac{7 + 1}{2},\dfrac{-3 + 9}{2} \right)$

$P=\left( \dfrac{8}{2},\dfrac{6}{2} \right)$

$P=\left( 4,3 \right)$

Slope of the line perpendicular to AB will be,

$m{{m}_{AB}}= -1$

$m \times  \left( -2 \right)= -1$

$m=\dfrac{1}{2}$

Equation of the perpendicular bisector of AB will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-3=\dfrac{1}{2}\left( x-4 \right)$

$2\left( y-3 \right)=\left( x-4 \right)$

$2y-6=x-4$

$2y=x+2$

(iii) Substitute x = -2 and y = p in the above equation.

$2p= -2+2$

$2p=0$

$p=0$

27. A and B are two points on the x-axis and y-axis respectively. P(2, -3) is the midpoint of AB. Find the 

(i) coordinates of A and B 

(ii) slope of line AB 

(iii) equation of line AB. 

Ans: (i) Let the coordinate of points A = (a, 0) and B = (0, b) respectively.

Point P is the midpoint. Therefore,

$P=\left( \dfrac{a + 0}{2},\dfrac{0 + b}{2} \right)$

$\left( 2, -3 \right)=\left( \dfrac{a}{2},\dfrac{b}{2} \right)$

Comparing both the sides,

$\dfrac{a}{2}=2$

$a=4$

And, $\dfrac{b}{2}= -3$

$b= -6$

Hence, the coordinate of points A = (4, 0) and B = (0, -6) respectively.

(ii) Slope of AB will be,

$m=\dfrac{-6 - 0}{0 - 4}$

$m=\dfrac{-6}{-4}$

$m=\dfrac{3}{2}$

(iii) Equation of the line AB will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-0=\dfrac{3}{2}\left( x-4 \right)$

$2y=3\left( x-4 \right)$

$2y=3x-12$

28. The equation of a line is 3x + 4y - 7 = 0. Find: 

(i) the slope of the line. 

(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x - y + 2 = 0 and 3x + y - 10 = 0.

Ans: (i) Given equation of the line is 3x + 4y - 7 = 0

Equate the given equation and compare with $y=mx+c$ to get the slope,

$3x+4y-7=0$

$4y= -3x+7$

$y= -\dfrac{3}{4}x+\dfrac{7}{4}$

Hence slope = ${{m}_{1}}= -\dfrac{3}{4}$

(ii) Slope of the line perpendicular to the given line is,

${{m}_{2}}{{m}_{1}}= -1$

${{m}_{2}}\times \left( -\dfrac{3}{4} \right)= -1$

${{m}_{2}}= \dfrac{4}{3}$

Given equations are,

$x-y+2=0$ ------ (1)

$3x+y-10=0$ ------ (2)

Adding equation (1) and (2),  we get

$x-y+2+3x+y-10=0$

$4x-8=0$

$4x=8$

$x=2$

The coordinate of the intersection point is (2, 4).

Equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-4=\dfrac{4}{3}\left( x-2 \right)$

$3\left( y-4 \right)=4\left( x-2 \right)$

$3y-12=4x-8$

$3y=4x+4$

29. ABCD is a parallelogram where A(x, y), B(5, 8), C(4, 7) and D(2, -4). Find: 

(i) Coordinates of A 

(ii) Equation of diagonal BD.

Ans: The given vertices of the parallelogram are A = (x, y), B = (5, 8), C = (4, 7) and D = (2, -4) 

In parallelogram, the diagonals bisect each other.

Therefore, the mid-point for diagonal AC will be the same for diagonal BD.

Thus, $\left( \dfrac{x + 4}{2},\dfrac{y + 7}{2} \right)=\left( \dfrac{5 + 2}{2},\dfrac{8 + \left( -4 \right)}{2} \right)$

$\left( \dfrac{x + 4}{2},\dfrac{y +  7}{2} \right)=\left( \dfrac{7}{2},2 \right)$

Comparing both the sides, we get

$\dfrac{x + 4}{2}=\dfrac{7}{2}$

$x+4=7$

$x=7-4$

$x=3$

And, $\dfrac{y + 7}{2}=2$

$y+7=4$

$y=4-7$

$y= -3$

Thus, the coordinate of A = (3, -3).

(ii) Slope of BD is,

$m=\dfrac{-4 - 8}{2 - 5}$

$m=\dfrac{-12}{-3}$

$m=4$

Equation of the diagonal BD will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-8=4\left( x-5 \right)$

$y=4x-20+8$

$y=4x-12$

30. Given equation of line ${{L}_{1}}$ is y = 4. 

(i) Write the slope of line ${{L}_{2}}$ if ${{L}_{2}}$ is the bisector of angle O. 

(ii) Write the coordinates of point P. 

(iii) Find the equation of ${{L}_{2}}$. 

Ans: (i) The line ${{L}_{2}}$ is the bisector of the angle O. Thus, the inclination of the line ${{L}_{2}}$ with the x-axis will be 45°.

Therefore slope of the line ${{L}_{2}}$ will be,

$m=tan \theta $

$m=tan {{45}^{o}}$

$m=1$

(ii) Given y = 4 and m = 1.

Slope of line ${{L}_{2}}$ is,

$m=\dfrac{y - 0}{x - 0}$

$1=\dfrac{4}{x}$

$x=4$

Hence, the coordinate of point P will be (4, 4).

(iii) Since, the line is passing through of the origin, therefore the equation of the line will be,

$y=mx+c$

$y=1$

31. Find:

(i) equation of AB 

(ii) equation of CD 

Ans: The given points are A = (-5, 4) and B = (3, 3)

(i) Slope of AB will be,

${{m}_{AB}}=\dfrac{3 - 4}{3 - \left( -5 \right)}$

${{m}_{AB}}=\dfrac{-1}{8}$

Equation of AB will be,

$y-{{y}_{1}}={{m}_{AB}}\left( x-{{x}_{1}} \right)$

$y-4= -\dfrac{1}{8}\left( x-\left( -5 \right) \right)$

$8\left( y-4 \right)= -\left( x+5 \right)$

$8y-32= -x-5$

$x+8y=27$

(ii) The coordinate of point D is D = (-3, 0)

Slope of CD will be,

${{m}_{CD}} {{m}_{AB}}= -1$

${{m}_{CD}} \times \left( \dfrac{-1}{8} \right)= -1$

${{m}_{CD}} =8$

Equation of CD will be,

$y-{{y}_{1}}={{m}_{CD}}\left( x-{{x}_{1}} \right)$

$y-0=8\left( x-\left( -3 \right) \right)$

$y=8x+24$

32. Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1. 

Ans: Given equation is $3x+5y=1$ and the coordinate of x-intercept is (-3, 0).

So the slope of the above equation is,

$3x+5y=1$

$5y= -3x+1$

$y= -\dfrac{3}{5}x+\dfrac{1}{5}$

Compare with equation $y=mx+c$

Hence slope = ${{m}_{1}}= -\dfrac{3}{5}$

Slope of the line perpendicular to the given line is,

${{m}_{1}}{{m}_{2}}= -1$

${{m}_{2}}\times \left( -\dfrac{3}{5} \right)= -1$

${{m}_{2}}=\dfrac{5}{3}$

Equation of the line will be,

$y-{{y}_{1}}={{m}_{2}}\left( x-{{x}_{1}} \right)$

$y-0=\dfrac{5}{3}\left( x-\left( -3 \right) \right)$

$3y=5\left( x+3 \right)$

$3y=5x+15$

33. A straight line passes through the points P(-1, 4) and Q(5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid-point of the line segment AB. Find: 

(i) the equation of the line. 

(ii) the coordinates of points A and B. 

(iii) the coordinates of point M. 

Ans: (i) The given points are P = (-1, 4) and Q = (5, -2)

Slope of PQ is,

$m=\dfrac{-2 - 4}{5 - \left( -1 \right)}$

$m=\dfrac{-6}{6}$

$m= -1$

Equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-4= -1\left( x-\left( -1 \right) \right)$

$y-4= -x-1$

$x+y=3$

(ii) Let the coordinates of points A and B are (x, 0) and (0, y) respectively.

Substitute y = 0 in the equation of the line,

$x+y=3$

$x+0=3$

$x=3$

Substitute x = 0 in the equation of the line,

$x+y=3$

$0+y=3$

$y=3$

Thus, the coordinates of points A = (3, 0) and B = (0, 3) respectively.

(iii) Point M is the midpoint of AB. Hence, the coordinate of point M will be,

$M=\left( \dfrac{3 + 0}{2},\dfrac{0 + 3}{2} \right)$

$M=\left( \dfrac{3}{2},\dfrac{3}{2} \right)$

34. In the given figure, line AB meets y-axis at point A. Line through C(2, 10) and D intersects line AB at right angle at point P. Find: 

(i) equation of line AB. 

(ii) equation of line CD. 

(iii) coordinates of points E and D. 

Ans: (i) Given points are A = (0, 6) and B = (-6, 8).

Slope of the AB is,

${{m}_{AB}}=\dfrac{8 - 6}{-6 - 0}$

${{m}_{AB}}=\dfrac{2}{-6}$

${{m}_{AB}}= -\dfrac{1}{3}$

Equation of AB will be,

$y-{{y}_{1}}={{m}_{AB}}\left( x-{{x}_{1}} \right)$

$y-6= -\dfrac{1}{3}\left( x-0 \right)$

$3\left( y-6 \right)= -x$

$x+3y=18$

(ii) Slope of CD will be,

${{m}_{CD}} {{m}_{AB}}= -1$

${{m}_{CD}}\left( -\dfrac{1}{3} \right)= -1$

${{m}_{CD}}=3$

The coordinate of point C = (2, 10).

Equation of CD will be,

$y-{{y}_{1}}={{m}_{CD}}\left( x-{{x}_{1}} \right)$

$y-10=3\left( x-2 \right)$

$y-10=3x-6$

$y=3x+4$

(iii) Let the coordinate of point E be $\left( {{x}_{1}},0 \right)$ and point D be $\left( {{x}_{2}},0 \right)$.

Substitute $x={{x}_{1}}$ and $y=0$ in equation AB.

$x+3y=18$

${{x}_{1}}+3\left( 0 \right)=18$

${{x}_{1}}=18$

Now, substitute $x={{x}_{2}}$ and $y=0$ in equation CD.

$y=3x+4$

$0=3{{x}_{2}}+4$

$3{{x}_{2}}= -4$

${{x}_{2}}= -\dfrac{4}{3}$

Thus, the coordinate of point E = (18, 0) and point D = $\left( -\dfrac{4}{3},0 \right)$.

35. A line through point P(4, 3) meets the x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.

Ans: Let the coordinate of points A is (a, 0) and B is (0, b).

Given P = (4, 3) and also point P divides AB such that BP : PA = 2 : 1.

Therefore,

$P=\left( \dfrac{2 \times  a + 1 \times  0}{2 + 1},\dfrac{2 \times  0 + 1 \times  b}{2 + 1} \right)$

$\left( 4,3 \right)=\left( \dfrac{2a}{3},\dfrac{b}{3} \right)$

Compare both the sides,

$\dfrac{2a}{3}=4$

$2a=12$

$a=\dfrac{12}{2}$

$a=6$

And, $\dfrac{b}{3}=3$

$b=9$

Hence the coordinates are A = (6, 0) and B = (0, 9)

Therefore slope of AB is,

$m=\dfrac{9 - 0}{0 - 6}$

$m=\dfrac{9}{-6}$

$m= -\dfrac{3}{2}$

Thus the equation of line AB is,

$y-{{y}_{1}}={{m}_{AB}}\left( x-{{x}_{1}} \right)$

$y-0= -\dfrac{3}{2}\left( x-6 \right)$

$2y= -3\left( x-6 \right)$

$2y= -3x+18$

$3x+2y=18$

36. Find the equation of line through the intersection of lines 2x - y = 1 and 3x + 2y = -9 and make an angle of 30° with positive direction of x-axis.

Ans: Given equations are,

$2x-y=1$ ------ (1)

And, $3x+2y= -9$ ------ (2)

Now, multiply equation (1) with 2 we get,

$4x-2y=2$ ------ (3)

Adding equation (2) and (3) we get,

$3x+2y+4x-2y= -9+2$

$7x= -7$

$x= -1$

Substitute value of x = -1 in equation (1) we get,

$2\left( -1 \right)-y=1$

$-2-y=1$

$-y=1+2$

$y= -3$

Thus, the coordinate of the intersection will be (-1, -3).

Slope of the line will be,

$m=tan {{30}^{o}}$

$m=\dfrac{1}{\sqrt{3}}$

Equation of the line will be,

$y-{{y}_{1}}={{m}_{{}}}\left( x-{{x}_{1}} \right)$

$y-\left( -3 \right)=\dfrac{1}{\sqrt{3}}\left( x-\left( -1 \right) \right)$

$\sqrt{3}\left( y+3 \right)=x+1$

$\sqrt{3}y+3\sqrt{3}=x+1$ ------ $\div \sqrt{3}$

$y=\dfrac{x}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{3\sqrt{3}}{\sqrt{3}}$

$y=\dfrac{x}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-3$

37. Find the equation of the line through the points A(-1, 3) and B(0, 2). Hence, show that the points A, B and C(1, 1) are collinear. 

Ans: Given points are A = (-1, 3) and B = (0, 2) and C = (1, 1)

Slope of AB is,

${{m}_{AB}}=\dfrac{2 - 3}{0 - \left( -1 \right)}$

${{m}_{AB}}=\dfrac{-1}{1}$

${{m}_{AB}}= -1$

Equation of the line passing through A and B will be,

$y-{{y}_{1}}={{m}_{AB}}\left( x-{{x}_{1}} \right)$

$y-3= -1\left( x-\left( -1 \right) \right)$

$y-3= -x-1$

$x+y=2$

Slope of BC will be,

${{m}_{BC}}=\dfrac{1 - 2}{1 - 0}$

${{m}_{BC}}=\dfrac{-1}{1}$

${{m}_{BC}}= -1$

Slope of AB = Slope of BC.

Hence, points A, B and C are collinear.

38. Three vertices of parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3, 2).

(i) the coordinate of the fourth vertex D

(ii) length of diagonal BD

(iii) equation of the side AD of the parallelogram ABCD

Ans: (i) Given vertices of the parallelogram are

A = (3, 6), B = (5, 10), C = (3, 2) and D = (x, y)

The diagonals of a parallelogram intersect each other.

As a result, the mid-point for diagonal AC and diagonal BD will be the same.

Equating for x-coordinate and y-coordinate using midpoint formula we get,

$\left( \dfrac{5 + x}{2},\dfrac{10 + y}{2} \right)=\left( \dfrac{3 + 3}{2},\dfrac{6 + 2}{2} \right)$

$\left( \dfrac{5 + x}{2},\dfrac{10 + y}{2} \right)=\left( \dfrac{6}{2},\dfrac{8}{2} \right)$

On comparing both the sides, we get

$\dfrac{5 + x}{2}=\dfrac{6}{2}$

$5+x=6$

$x=1$

And, $\dfrac{10 + y}{2}=\dfrac{8}{2}$

$10+y=8$

$y= -2$

Thus, the coordinate of D will be (1, -2)

(ii) Length of diagonal BD can be calculated by using distance formula

$BD=\sqrt{{{(1-5)}^{2}}+{{(-2-10)}^{2}}}$

$BD=\sqrt{{{(-4)}^{2}}+{{(-12)}^{2}}}$

$BD=\sqrt{16+144}$

$BD=\sqrt{160}$

$BD=4\sqrt{10}  $units

(iii) Slope of AB is,

$m=\dfrac{10 - 6}{5 - 3}$

$m=\dfrac{4}{2}$

$m=2$

Equation of side AB will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-6=2\left( x-3 \right)$

$y-6=2x-6$

$y=2x$

39. In the given figure ABC is a triangle and BC is parallel to the y-axis. AB and AC intersects the y-axis at P and Q respectively.

(i) Write the coordinates of A.

(ii) Find the length of AB and AC.

(iii) Find the ratio in which Q divides AC.

(iv) Find the equation of the line AC.

Ans: 

(i) The coordinate of point A is (4,0).

(ii) Coordinate of points B = (-2, 3) and C = (-2, -4)

Length of AB and AC by using distance formula is,

$AB=\sqrt{{{(4-\left( -2 \right))}^{2}}+{{(0-3)}^{2}}}$

$AB=\sqrt{{{(6)}^{2}}+{{(-3)}^{2}}}$

$AB=\sqrt{36+9}$

$AB=\sqrt{45}$

$AB=3\sqrt{5}$units

And,

$AC=\sqrt{{{(4-\left( -2 \right))}^{2}}+{{(0-\left( -4 \right))}^{2}}}$

$AC=\sqrt{{{(6)}^{2}}+{{(4)}^{2}}}$

$AC=\sqrt{36+16}$

$AC=\sqrt{52}$

$AC=2\sqrt{13}$ units

(iii) Slope of the AC is,

${{m}_{AC}}=\dfrac{-4 - 0}{-2 - 4}$

${{m}_{AC}}=\dfrac{-4}{-6}$

${{m}_{AC}}=\dfrac{2}{3}$

Equation of AC will be,

$y-{{y}_{1}}={{m}_{AC}}\left( x-{{x}_{1}} \right)$

$y-0=\dfrac{2}{3}\left( x-4 \right)$

$3y=2\left( x-4 \right)$

$3y=2x-8$

Point Q lies on the y-axis.

Thus, the coordinate of point Q is (0, y).

Now, substitute x = 0 in the equation of AC.

$3y=2x-8$

$3y=2\left( 0 \right)-8$

$3y= -8$

$y= -\dfrac{8}{3}$

Thus, the coordinate of point Q = $\left( 0,-\dfrac{8}{3} \right)$

Let the Q divides the AC in k :1, thus,

$Q=\left( \dfrac{m{{x}_{2}} + n{{x}_{1}}}{m + n},\dfrac{m{{y}_{2}} + n{{y}_{1}}}{m + n} \right)$

$\left( 0,-\dfrac{8}{3} \right)=\left( \dfrac{k \times  \left( -2 \right) + 1 \times  4}{k + 1},\dfrac{k \times  \left( -4 \right) + 1 \times  0}{k + 1} \right)$

$\left( 0,-\dfrac{8}{3} \right)=\left( \dfrac{-2k + 4}{k + 1},\dfrac{-4k}{k + 1} \right)$

Comparing both the sides, we get

$\dfrac{-2k + 4}{k + 1}=0$

$-2k+4=0$

$-2k= -4$

$k=\dfrac{4}{2}$

$k=2$

Thus, the required ration will be 2 : 1

(iv) Equation of line AC is $3y=2x-8$

$2x-3y=8$

40. The slope of a line joining P(6, k) and Q(1-3k, 3) is $\dfrac{1}{2}$. Find:

(i) k

(ii) Midpoint of PQ, using the value of ‘k’ found in (i)

Ans: Given points are P = (6, k) and Q = (1-3k, 3) and $m=\dfrac{1}{2}$

Slope of PQ is,

$m=\dfrac{3 - k}{1 - 3k - 6}$

$\dfrac{1}{2}=\dfrac{3 - k}{1 - 3k - 6}$

$1-3k-6=2\left( 3-k \right)$

$1-3k-6=6-2k$

$-3k+2k=6+6-1$

$k=-11$

(ii) The coordinate of P will be (6, -11).

Substitute k value in the Q coordinates,

Coordinate of $Q=\left( 1-3k,3 \right)$

$Q=\left( 1-3\left( -11 \right),3 \right)$

$Q=\left( 1+33,3 \right)$

$Q=\left( 34,3 \right)$

Coordinate of midpoint will be,

$A=\left( \dfrac{6 + 34}{2},\dfrac{-11 + 3}{2} \right)$

$A=\left( \dfrac{40}{2},\dfrac{-8}{2} \right)$

$A=\left( 20,-4 \right)$

41. A line AB meets X-axis at A and Y-axis at B. P(4, -1) divides AB in the ratio 1 : 2. 

(i) Find the coordinates of A and B. 

(ii) Find the equation of the line through P and perpendicular to AB.

Ans: (i) The coordinate of points A and B are (a, 0) and (0, b)

Point P divides AB in a ratio of 1 : 2.

$P=\left( \dfrac{2 \times  a + 1 \times  0}{1 + 2},\dfrac{2 \times  0 + 1 \times  b}{1 + 2} \right)$

$\left( 4,-1 \right)=\left( \dfrac{2a}{3},\dfrac{b}{3} \right)$

Comparing both the sides, we get

$\dfrac{2a}{3}=4$

$2a=12$

$a=\dfrac{12}{2}$

$a=6$

And, 

$\dfrac{b}{3}= -1$

$b= -3$

Thus the coordinate of points A = (6, 0) and B = (0, -3) respectively.

(ii) Slope of AB is,

${{m}_{AB}}=\dfrac{-3 - 0}{0 - 6}$

${{m}_{AB}}=\dfrac{-3}{-6}$

${{m}_{AB}}=\dfrac{1}{2}$

Slope of line perpendicular to AB will be,

$m{{m}_{AB}}= -1$

$m\times \left( \dfrac{1}{2} \right)= -1$

$m= -2$

Equation of the line will be,

$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-\left( -1 \right)= -2\left( x-4 \right)$

$y+1= -2x+8$

$2x+y=7$.

ICSE Class 10 Mathematics Chapter 14 Selina Concise Solutions

Introduction

This is one of the important chapters for class 10 ICSE. The topics are the fundamentals of the straight-line equation, the inclination of a line, the concept of slope (gradient), the slope of a straight line passing through two given fixed points, slopes of parallel and straight lines, various forms of the equations of straight lines and finding x/y -intercepts of a given line. Students should understand the concepts of these topics so that they have a strong foundation when they go to class 11 and class 12. Students can also refer to Vedantu’s NCERT solutions to have a good grip of the topics and practice with the exercise given. 

Maths is an interesting subject and can be mastered only through practice. So make a routine and practice every day which will help you to reduce the silly mistakes that often students do and lose marks unnecessarily.

Tips for preparation for ICSE Class 10 Maths Chapter 14

Find some of the methods below to learn the chapter in an easier way:

1. This chapter has lots of formulas so to make it easier for the revision make a list of the formulas from the particular chapter. Every day before starting your studies just go through the formulas. When you keep reading every day it automatically captures your mind.

2. While learning the formulas understand where it is used and also the derivation has to be learned so that in any circumstances you happen to forget you can always derive the same.

3. Always clarify the doubts if you have any while preparing. Try not to leave it pending as you might forget and this will remain as it is.

4. NCERT books have to be well studied and practiced with focus and concentration.

5. Practice solving problems with different concepts.

6. Complete all the exercises in NCERT; try not to skip anything while practicing. All the problems should be solved in the notes.

7. When you start with the new chapter just solve the few problems from the previous chapter. This will help you to remember and give better practice.

8. Solve the problems in steps. This will help you to reduce the mistakes and if you go wrong it is easier to rectify.

9. All the syllabus should be completed before time allowing some time to revise before the exam.

10.  Solve the problems by paying attention to decimals and the signs.

11.  Practice answering the previous year’s question paper. Students should solve at least the previous 5-year question paper to have an accurate practice of the chapter.

12.  Pay attention and recognize the trends of the questions in the question papers.

13.  Know your strengths and weaknesses and recognize the areas where you can answer the questions easily and which part you had to struggle with. Knowing these points you can improve and rectify the common mistakes.

14.  Focus on increasing the speed while answering so you can complete the question paper in the given time and have sufficient time for revision.

15. Create a schedule and focus on the weak points and also practice the areas which you are good at.

16. Understand the applications with theorems and the concepts.

17. Read the ist of the formulas at least four times a day.

18. Practice more and more to ace the Maths exam.

19. Be clear with the concepts.

20. Try not to skip any chapters or topics. Practice and know the various questions and exact methods to answer them by going through the solutions provided by Vedantu.