## Crucial Practice Problems for CBSE Class 11 Maths Chapter 10: Straight Lines

Important Questions for **Class 11 Maths Chapter 10 - Straight Lines** are available in Vedantu. These Questions are designed as per the latest Syllabus of NCERT Curriculum and are frequently asked in exams in different types. Solving these important questions will definitely help students to score good marks in final exams. All the solutions are explained in a detailed manner along with the graphical representation. Students can refer to these solutions for learning the most important questions and prepare for their board exams.

## Download CBSE Class 11 Maths Important Questions 2024-25 PDF

Also, check CBSE Class 11 Maths Important Questions for other chapters:

CBSE Class 11 Maths Important Questions | ||

Sl.No | Chapter No | Chapter Name |

1 | Chapter 1 | |

2 | Chapter 2 | |

3 | Chapter 3 | |

4 | Chapter 4 | |

5 | Chapter 5 | |

6 | Chapter 6 | |

7 | Chapter 7 | |

8 | Chapter 8 | |

9 | Chapter 9 | |

10 | Chapter 10 | Straight Lines |

11 | Chapter 11 | |

12 | Chapter 12 | |

13 | Chapter 13 | |

14 | Chapter 14 | |

15 | Chapter 15 | |

16 | Chapter 16 |

## Topics Covered in Class 11 Maths Chapter 10 - Straight Lines are as follows:

Slope of a Line

Slope of a line when coordinates of any two points on the line are given

Conditions for parallelism and perpendicularity of lines in terms of their slopes

Angle between two lines

Collinearity of three points

Various Forms of the Equation of a Line

Horizontal and vertical lines

Point-slope form

Two-point form

Slope-intercept form

Intercept – form

Normal form

General Equation of a Line

Different forms of Ax + By + C = 0f

Distance of a Point From a Line

Distance between two parallel lines

## Study Important Questions for Class 11 Mathematics Chapter 10 - Straight Lines

Very Short Answer Questions. (1 Mark)

1. Find the slope of the lines passing through the point $\left( {3,\, - 2} \right)$ and $\left( { - 1,\,4} \right)$

Ans: Slope of the line = $m$

The given points are $\left( {3,\, - 2} \right)$ and $\left( { - 1,\,4} \right)$.

Let $\left( {{x_1},\,\,{y_1}} \right)$ be $\left( {3,\, - 2} \right)$ and $\left( {{x_2},\,\,{y_2}} \right)$ be $\left( { - 1,\,4} \right)$.

$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

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

$\Rightarrow \dfrac{6}{{ - 4}} = - \dfrac{3}{2}$

Therefore, the slope of the line passing through the point $\left( {3,\, - 2} \right)$ and $\left( { - 1,\,4} \right)$ is $ - \dfrac{3}{2}$.

2. Three points $P\left( {h,\,\,k} \right),\,\,Q\left( {{x_1},\,\,{y_1}} \right)$ and $R\left( {{x_2},\,\,{y_2}} \right)$ lie on a line. Show that

$\left( {h - {x_1}} \right)\left( {{y_2} - {y_1}} \right) = \left( {k - {y_1}} \right)\left( {{x_2} - {x_1}} \right)$

Ans: The given points are $P\left( {h,\,\,k} \right),\,\,Q\left( {{x_1},\,\,{y_1}} \right)$and $R\left( {{x_2},\,\,{y_2}} \right)$.

Since, the points $P,\,\,Q$ and $R$ lie on a line. Therefore, they are collinear.

Hence,

Slope of $PQ$= Slope of $QR$

$ \Rightarrow \dfrac{{{y_1} - k}}{{{x_1} - h}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ $\left[ {\because \,\,{\text{Slope }} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right]$

$ \Rightarrow \dfrac{{\left( {k - {y_1}} \right)}}{{\left( {h - {x_1}} \right)}} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ (On cross multiplication)

$\left( {h - {x_1}} \right)\left( {{y_2} - {y_1}} \right) = \left( {k - {y_1}} \right)\left( {{x_2} - {x_1}} \right)$

3. Write the equation of the line through the points $\left( {1, - 1} \right)$ and \[\left( {3,\,\,5} \right)\]

Ans: We know that equation of line through two points $\left(x_{1}, y_{1}\right) \&\left(x_{2}, y_{2}\right)$ is

$y-y_{1}=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$

Since the equation of the line through the points $(1,-1)$ and $(3,5)$.

Here,

$x_{1}=1, y_{1}=-1$

$x_{2}=3 \text { and } y_{2}=5$

Putting values

$(y-(-1))=\dfrac{5-(-1)}{3-1}(x-1)$

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

$y+1=\dfrac{6}{2}(x-1)$

$y+1=3(x-1)$

$y+1=3 x-3$

$y-3 x+1+3=0$

$y-3 x+4=0$

Hence the required equation is $\mathrm{y}-3 \mathrm{x}+4=0$

4. Find the measure of the angle between the lines $x + y + 7 = 0$ and $x - y + 1 = 0$.

Ans: The given equation of lines are $x + y + 7 = 0$ and $x - y + 1 = 0$.

Express the given equation as slope-intercept form $y = mx + c$

where,

(Slope)$m$= coefficient of $x$

$\mathrm{x}+\mathrm{y}+7=0$

$\mathrm{~m}_{1}=\dfrac{-1}{1}$

$\mathrm{x}-\mathrm{y}+1=0$

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

Product of these two slopes is $-1$, therefore, the lines are at right angles.

5. Find the equation of the line that has $y - $intercept $4$ and is perpendicular to the line $y = 3x - 2$.

Ans: The given equation of the is $y = 3x - 2$.

Express the given equation as slope-intercept form $y = mx + c$

where,

(Slope)$m$= coefficient of $x$

${m_1} = 3$

When the lines are perpendicular. Then the product of slope is $ - 1$.

$\therefore \,\,\,{m_1}.{m_2} = - 1$

$\,\,\,\,\,\,\,\,\,\,3.{m_2} = - 1$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,{m_2} = - \dfrac{1}{3}$

Given,

$y - $intercept of other line is 4.

Therefore, the required equation of the line using the slope-intercept form $y = mx + c$.

$y = - \dfrac{1}{3}x + 4$

6. Find the equation of the line, which makes intercepts $ - 3$ and 2 on the $x$ and $y$-axis respectively.

Ans: Given,

($x - $intercept) a $ = - 3$

($y - $intercept) b $ = 2$

Required equation is given by $\dfrac{x}{a} + \dfrac{y}{b} = 1$

$a = - 3,b = 2$

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

$2x - 3y + 6 = 0$

7. Equation of a line is $3x - 4y + 10 = 0$ find its slope.

Ans: Given,

Equation of the line is $3x - 4y + 10 = 0$.

Slope of the line is given by $m = \dfrac{{ - \,{\text{co - efficient of }}x}}{{{\text{co - efficient of }}y}}$

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

8. Find the distance between the parallel lines $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$.

Ans:Given,

Equations of the parallel lines are $3x - 4y + 7 = 0$ and $3x - 4y + 5 = 0$.

The general equation of the parallel lines is given by $Ax + By + {C_1} = 0{\text{ and }}Ax + By + {C_2} = 0$.

On comparing the given equation of two parallel lines with general equation, we get

$A = 3,\,\,B = - 4,\,{C_1} = 7\,{\text{and }}{C_2} = 5$.

Distance between two parallel lines is $d = \dfrac{{\left| {{C_1} - {C_2}} \right|}}{{\sqrt {{A^2} + {B^2}} }}$

$\Rightarrow \dfrac{{|7 - 5|}}{{\sqrt {{{(3)}^2} + {{( - 4)}^2}} }}$

$\Rightarrow \dfrac{2}{{\sqrt {9 + 16} }}$

$\Rightarrow \dfrac{2}{5}$

9. Find the equation of a straight line parallel to the $y$-axis and passing through the point $\left( {4,\, - 2} \right)$.

Ans: Given point is $\left( {4,\, - 2} \right)$

Equation of line parallel to $y$-axis is $x = a\,\,\,\,\,...(i)$

Since, equation (i) passing through $\left( {4,\, - 2} \right)$

$\therefore \,\,a = 4$

So,

$x = 4$

$x - 4 = 0$

10. If $3x - by + 2 = 0$ and $9x + 3y + a = 0$ represent the same straight line, find the values of $a$ and $b$.

Ans: Given the equations of the line are $3x - by + 2 = 0$ and $9x + 3y + a = 0$.

According to the question,

$\dfrac{3}{9} = \dfrac{{ - b}}{3} = \dfrac{2}{a}$

On comparing the values, we get

$\dfrac{3}{9} = \dfrac{{ - b}}{3}\,\,\,\,\,{\text{and }}\dfrac{3}{9} = \dfrac{2}{a}$

$\therefore \,\,b = - 1\,\,{\text{and }}\,a = 6{\text{ }}$

11. Find the distance between $P\left( {{x_1}{y_1}} \right)$ and $Q\left( {{x_2},{y_2}} \right)$ when ${\mathbf{PQ}}$ is parallel to the $y$-axis.

Ans. When PQ is parallel to the $y$-axis,

Then ${x_1} = {x_2}$

$\therefore PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

$= \sqrt {{{\left( {{x_2} - {x_2}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$

$= \left| {{y_2} - {y_1}} \right|$

12. Find the slope of the line, which makes an angle of ${30^\circ }$ with the positive direction of $y - $axis measured anticlockwise.

Ans: Given,

Line which makes an angle of ${30^\circ }$ with the positive direction of $y - $axis measured anticlockwise.

Let $\theta $ be the inclination of the line,

$\theta = {120^\circ }$

(slope) $m = \tan 120^\circ $

$= \tan \left( {90 + 30} \right)$

$= - \sqrt 3$

13. Determine $x$ so that the inclination of the line containing the points $(x, - 3)$ and $(2,5)$ is ${135^\circ }$.

Ans: We have been given a line joining the points $(x,-3)$ and $(2,5)$ which makes an angle of $135^{\circ}$ with the $x$-axis.

We know that the slope of a line joining the two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is equal to the tangent of the angle made by the line with $x$-axis in anticlockwise direction given by as follows: slope $=\tan \theta=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

So we have $\theta=135^{\circ}, x_{1}=2, x_{2}=x, y_{1}=5, y_{2}=-3$

$\Rightarrow \tan 135^{\circ}=\dfrac{-3-5}{x-2}$

Since we know that $\tan 135^{\circ}=\tan \left(90^{\circ}+45^{\circ}\right)=-\cot 45^{\circ}=-1$ as in second quadrant tangent function is negative.

$\Rightarrow-1=\dfrac{-3-5}{x-2}$

$\Rightarrow-1=\dfrac{-8}{x-2}$

On cross multiplication, we get as follows:

$\Rightarrow-x+2=-8$

On adding $-2$ to both the sides of the equation, we get as follows:

$\Rightarrow-x-2+2=-8-2$

$\Rightarrow-x=-10$

$\Rightarrow x=10$

Therefore the value of $\mathrm{x}$ is equal to 10 .

14. Find the distance of the point $(4,1)$ from the line $3x - 4y - 9 = 0$

Ans: Given,

Point $(4,1)$ and equation of the line $3x - 4y - 9 = 0$

Let ${\text{d}}$ be the required distance,

$d = \dfrac{{|3 \cdot (4) - 4(1) - 9|}}{{\sqrt {{{(3)}^2} + {{( - 4)}^2}} }}$

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

15. Find the value of $x$ for which the points $(x, - 1),(2,1)$ and $(4,5)$ are collinear.

Ans. Given three collinear points.

Let the point be $A(x, - 1),B(2,1),C(4,5)$

Since, the points are collinear. Therefore,

Slope of ${\text{AB}} = $ Slope of ${\text{BC}}$

$\dfrac{{1 + 1}}{{2 - x}} = \dfrac{{5 - 1}}{{4 - 2}}$

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

$2 - x = 1$

$ - x = - 1$

$x = 1$

16. Find the angle between the $x$-axis and the line joining the points $(3, - 1)$ and $\left( {4,\, - 2} \right)$.

Ans: Given,

Line joining the points $(3, - 1)$ and $\left( {4,\, - 2} \right)$.

${m_1} = 0\quad $ Slope of $x$-axis

${m_2} = $ slope of line joining points $(3, - 1)$ and $(4, - 2)$

${m_2} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

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

Angle between two line is given by $\tan \theta = \left| {\dfrac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|$

$\tan \theta = \left| {\dfrac{{-1 + 0}}{{1 + 0 \times ( - 1)}}} \right|$

$\tan \theta = 1$

$\theta = {45^\circ }$

17. Using slopes, find the value of $x$ for which the points $(x, - 1),\,(2,1)$ and $(4,5)$ are collinear.

Ans: Given,

Three collinear points.

Let the points be A$\left( {x,\, - 1} \right)$, B$\left( {2,\,\,1} \right)$ and C $\left( {4,\,\,5} \right)$.

Since, the points are collinear. Therefore,

Slope of ${\text{AB}} = $ Slope of ${\text{BC}}$

$\Rightarrow \dfrac{{1 - \left( { - 1} \right)}}{{2 - x}} = \dfrac{{5 - 1}}{{4 - 2}}$

$\Rightarrow \,\,\,\,\,\,\,\,\,\dfrac{{1 + 1}}{{2 - x}} = \dfrac{4}{2} $

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2 = 2\left( {2 - x} \right) $

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 = 2 - x $

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1$

18. Find the value of $K$ so that the line $2x + ky - 9 = 0$ may be parallel to $3x - 4y + 7 = 0$

Ans: Given,

The Equations of two parallel lines are $2x + ky - 9 = 0$ and $3x - 4y + 7 = 0$.

Slope of the line is given by $m = \dfrac{{ - \,{\text{co - efficient of }}x}}{{{\text{co - efficient of }}y}}$

According to the question,

Slope of 1st line = slope of 2nd line

$\dfrac{{ - 2}}{k} = \dfrac{{ - 3}}{{ - 4}}$

$ \Rightarrow k = \dfrac{{ - 8}}{3}$

19. Find the value of $K$, given that the distance of the point $(4,1)$ from the line $3x - 4y + K = 0$ is 4 units.

Ans: Given,

The distance of the point $(4,1)$ from the line $3x - 4y + k = 0$ is 4 units.

$\Rightarrow \dfrac{{|3(4) - 4(1) + k|}}{{\sqrt {{{(3)}^2} + {{( - 4)}^2}} }} = 4 $

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\dfrac{{\left| {12 - 4 + k} \right|}}{{\sqrt {25} }} = 4 $

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{{\left| {8 + k} \right|}}{5} = 4 $

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left| {8 + k} \right| = 20$

When,

$8 + k = 20$

$\,\,\,\,\,\,\,k = 12$

When,

$- \left( {8 + k} \right) = 20$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k = - 28$

20. Find the equation of the line through the intersection of $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$ which cuts off equal intercepts on the axes.

Ans: Given,

Equation of two intersecting lines $3x - 4y + 1 = 0$ and $5x + y - 1 = 0$.

Slope of a line which makes equal intercept on the axes is $ - 1$ any line through the intersection of given lines is,

$(3x - 4y + 1) + K(5x - y - 1) = 0$

$(3 + 5K)x + y(K - 4) + 1 - K = 0$

Slope of the line is given by $m = \dfrac{{ - \,{\text{co - efficient of }}x}}{{{\text{co - efficient of }}y}}$

$\Rightarrow \,\,\,\,m = - \dfrac{{(3 + 5K)}}{{K - 4}}$

$\Rightarrow - 1 = - \dfrac{{(3 + 5K)}}{{K - 4}}$

$\Rightarrow \,\,\,K = \dfrac{{ - 7}}{4}$

21. Find the distance of the point $(-2,\,\,3)$ from the line $12x - 5y = 2$.

Ans: Given point $(-2,\,\,3)$ and equation of the line $12x - 5y = 2$.

Since, we know

Distance of a point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ from the line $\mathrm{ax}+\mathrm{by}+\mathrm{c}=0$ is $\dfrac{\left|\mathrm{ax}_{1}+\mathrm{by}_{1}+\mathrm{c}\right|}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}$ Distance of the point $(-2,3)$ from the line $12 \mathrm{x}-5 \mathrm{y}-2=0$ is

$\dfrac{|12(-2)-5(3)-2|}{\sqrt{12^{2}+(-5)^{2}}}=\dfrac{|-24-15-2|}{13}=\dfrac{41}{13}$

22. Find the equation of a line whose perpendicular distance from the origin is 5 units and angle between the positive direction of the $x$-axis and the perpendicular is ${30^\circ }$.

Ans. Given,

Perpendicular distance of the line from the origin is 5 units.

Angle between the positive direction of the $x$-axis and the perpendicular is ${30^\circ }$.

Hence,

$p = 5,\alpha = {30^\circ }$

Required equation is given by $x\cos \alpha + y\sin \alpha = p$.

$x\cos {30^\circ } + y\sin {30^\circ } = 5$

$\sqrt 3 x + y - 10 = 0$

23. Write the equation of the lines for which $\tan \theta = \dfrac{1}{2}$, where ${\mathbf{Q}}$ is the inclination of the line and $x$ intercept is 4.

Ans. Given,

Slope of the line will become;

$m = \tan \theta = \dfrac{1}{2}$ and

$x - $intercept, i.e., d = 4

$y = \dfrac{1}{2}(x - 4)\,\,\,\,\,\,\,\,\,\left[ {\,\because y = m(x - d)} \right]$

$2y - x + 4 = 0$

24. Find the angle between the $x$-axis and the line joining the points $(3, - 1)$ and $(4, - 2)$.

Ans: Given,

Two distinct points.

Let $A(3, - 1)$ and $B(4, - 2)$

Slope of AB,

$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

$\,\,\,\,\, = \dfrac{{ - 2 - ( - 1)}}{{4 - 3}}$

$\,\,\,\,\, = - 1$

$\,\,\,\,\,\,\because m = \tan \theta$

$\therefore \tan \theta = - 1$

$\theta=135$

where $\theta $ is the angle which AB makes with the positive direction of the x axis.

25. Find the equation of the line intersecting the $x$-axis at a distance of 3 units to the left of origin with slope $ - 2$.

Ans. Given,

Line intersecting the $x$-axis at a distance of 3 units to the left of origin.

Therefore, the point is $\left( { - 3,\,0} \right)$

Slope of the line = m $ = - 2$

Required equation is given by one-point form,

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y - {y_1} = m\left( {x - {x_1}} \right)$

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,y - 0 = - 2(x + 3)$

$\Rightarrow 2x + y + 6 = 0$

Short Answer Question (4 Marks)

1. If $p$ is the length of the $ \bot $ from the origin on the line whose intercepts on the axes are a and b. Show that $\dfrac{1}{{{p^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}$

Ans: Equation of the line is $\dfrac{x}{a} + \dfrac{y}{b} = 1$

$ \Rightarrow \dfrac{x}{a} + \dfrac{y}{b} - 1 = 0$

The distance of this line from the origin is P

$\therefore P = \dfrac{{\left| {\dfrac{0}{a} + \dfrac{0}{b} - 1} \right|}}{{\sqrt {{{\left( {\dfrac{1}{a}} \right)}^2} + {{\left( {\dfrac{1}{b}} \right)}^2}} }}\quad \left[ {d = \dfrac{{|ax + by + c|}}{{\sqrt {{a^2} + {b^2}} }}} \right]$

$\dfrac{P}{1} = \dfrac{1}{{\sqrt {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} }}$

$\dfrac{1}{P} = \sqrt {\dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}} $

Squaring both the sides,

$\dfrac{1}{{{P^2}}} = \dfrac{1}{{{a^2}}} + \dfrac{1}{{{b^2}}}$

Hence proved.

2. Find the value of $p$ so that the three lines $3x + y - 2 = 0,px + 2y - 3 = 0$ and $2x - y - 3 = 0$ may intersect at one point.

Ans: Given,

Equation of three lines,

$3x + y - 2 = 0\,\,\,\,\,\,\,.....(i)$

$px + 2y - 3 = 0\,\,\,\,\,.....(ii)$

$2x - y + 3 = 0\,\,\,\,\,\,\,\,.....(iii)$

On solving equation (i) and (iii),

$x = 1$ and $y = - 1$

Put x, y in equation (ii),

$\,\,\,\,\,\,\,\,\,\,px + 2y - 3 = 0$

$p.1 + 2.\left( { - 1} \right) - 3 = 0$

$p - 2 - 3 = 0$

$\,\,\,\,\,\,\,\,p - 5 = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,p = 5$

3. Find the equation to the straight line which passes through the point $(3,4)$ and has an intercept on the axes equal in magnitude but opposite in sign.

Ans: Given the makes equal intercepts in axes but opposite in sign.

Let the intercepts be,

($x - $intercept) a $ = a$

($y - $intercept) b $ = b$

The equation of the line is given by $\dfrac{x}{a} + \dfrac{y}{b} = 1$.

On substituting the values,

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

$ \Rightarrow x - y = a \ldots \ldots .(i)$

Since it passes through the point $(3,4)$

$3 - 4 = a$

$a = - 1$

Put the value of a in equation (i)

$x - y = - 1$

$x - y + 1 = 0$

4. By using the area of $\Delta .$ Show that the points $(a,b + c),\,\,(b,c + a)$ and $(c,a + b)$ are collinear.

Ans: Given,

Three points, let the points be ${\text{A}}(a,b + c),\,\,{\text{B}}(b,c + a)$ and ${\text{C}}(c,a + b)$

We know that

Area of triangle is given by

$\Delta=\dfrac{1}{2} \begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{vmatrix}$

Here,

$x_{1}=a, y_{1}=b+c$

$x_{2}=b, y_{2}=c+a$

$x_{3}=c, y_{3}=a+b$

Putting values

$\Delta=\dfrac{1}{2} \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1\end{vmatrix}$

Applying $C_{1} \rightarrow C_{1}+C_{2}$

$\Delta=\dfrac{1}{2} \begin{vmatrix} a+b+c & b+c & 1 \\ b+c+a & c+a & 1 \\ c+a+b & a+b & 1\end{vmatrix}$

Taking $(a+b+c)$ common from $C_{1}$

$\Delta=\dfrac{1}{2}(a+b+c) \begin{vmatrix} 1 & b+c & 1 \\ 1 & c+a & 1 \\ 1 & a+b & 1\end{vmatrix}$

Here, $1^{\text {st }}$ and $3^{\text {rd }}$ Column are Identical

Hence value of determinant is zero

$\Delta=\dfrac{1}{2}(a+b+c) \times 0 $

$\Delta=0$

So, $\Delta=\mathbf{0}$

Hence points $A, B \& C$ are collinear

5. Find the slope of a line, which passes through the origin, and the midpoint of the line segment joining the point $P(0, - 4)$ and $Q(8,0)$

Ans: Given,

Line segment joining the point $P(0, - 4)$ and $Q(8,0)$.

Let $M$ be the midpoint of segment PQ then

$M = \left( {\dfrac{{{x_1} + {x_2}}}{2} + \dfrac{{{y_1} + {y_2}}}{2}} \right)$

$\,\,\,\,\,\,\,\,\,\, = \left( {\dfrac{{0 + 8}}{2},\dfrac{{ - 4 + 0}}{2}} \right)$

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

Slope of line joining origin $O\left( {0,\,0} \right)$ and the mid-point $M\left( {4, - 2} \right)$,

$OM = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$

$\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - 2 - 0}}{{4 - 0}}$

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

6. Find the equation of the line passing through the point $(2,2)$ and cutting off intercepts on the axes whose sum is 9.

Ans: Required equation is $\dfrac{x}{a} + \dfrac{y}{b} = 1 \ldots \ldots (i)$

Where,

($x - $intercept) $ = a$

($y - $intercept) $ = b$

Given,

Sum of intercepts = 9

$\therefore \,\,a + b = 9$

$b = 9 - a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( {ii} \right)$

Substituting the value of equation (ii) in equation (i),

$ \Rightarrow \dfrac{x}{a} + \dfrac{y}{{9 - a}} = 1$

This line passes through $(2,2)$

$\therefore \dfrac{2}{a} + \dfrac{2}{{9 - a}} = 1$

${a^2} - 9a + 18 = 0$

${a^2} - 6a - 3a + 18 = 0$

$a(a - 6) - 3(a - 6) = 0$

$(a - 6)(a - 3) = 0$

$a = 6,3$

$a = 6\,\,\,{\text{or}}\,\,\,a = 3$

On substituting the values of a in equation (ii), we get

$b = 3\,\,\,\,{\text{or }}\,\,b = 6$

Therefore, the equation of line are,

$\dfrac{x}{6} + \dfrac{y}{3} = 1\quad \dfrac{x}{3} + \dfrac{y}{6} = 1$

7. Reduce the equation $\sqrt 3 x + y - 8 = 0$ into normal form. Find the values $p$ and $\omega $.

Ans: Given,

Equation of a line$\sqrt 3 x + y - 8 = 0$

Let $\sqrt 3 x + y = 8 \ldots \ldots .(i)$

$\sqrt {{{(\sqrt 3 )}^2} + {{(1)}^2}} = 2$

Dividing (i) by 2

$\dfrac{{\sqrt 3 }}{2}x + \dfrac{y}{2} = 4$

$x\cos {30^\circ } + y \cdot \sin {30^\circ } = 4 \ldots \ldots .(ii)$

On comparing the above equation with the standard from,

$x\cos \omega + y\sin \omega = p$

Where,

$p$is the perpendicular distance from the origin

$\omega $ is the angle between perpendicular and the positive $x - $axis

$p = 4 $

$\omega = \dfrac{\pi }{6}$

8. Without using the Pythagoras theorem show that the points $(4,4),(3,5)$ and $( - 1, - 1)$ are the vertices of a right angled $\Delta $.

Ans: Let given points be $A(4,4),B(3,5)$ and $C( - 1, - 1)$

Slope of AB,

${m_1} = \dfrac{{5 - 4}}{{3 - 4}} = - 1$

Slope of BC,

${m_2} = \dfrac{{ - 1 - 5}}{{ - 1 - 3}}$

$\Rightarrow \dfrac{{ - 6}}{{ - 4}} = \dfrac{3}{2}$

Slope of AC,

${m_3} = \dfrac{{ - 1 - 4}}{{ - 1 - 4}} = + 1$

Since,

Slope of $AB \times $ slope of $AC = - 1$

${m_1} \times {m_3} = - 1$

$ \Rightarrow AB \bot AC$ (if the product of slope is $ - 1$, then the lines are perpendicular)

Hence $\Delta {\text{ABC}}$ is right angled at ${\text{A}}$.

9. The owner of a milk store finds that he can sell 980 litres of milk each week at Rs 14 per litre and 1220 litre of milk each week at Rs 16 per litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17 per litre?

Ans. Let selling price be P along $x$-axis

demand of milk be D along $y$-axis

We know that the equation of line is

$y=m x+c$

Here, $P$ is along $x$-axis and $D$ is along $y$-axis

So, our equation becomes

$D=m P+c$

Now,

Owner sells 980 litre milk at Rs 14 /litre

So, $D=980 \& P=14$ satisfies the equation

Putting values in (1)

$980=14 m+c$

Owner sells 1220 litre milk at Rs $16 /$ litre

So, $D=1220 \& P=16$ satisfies the equation

Putting values in (1)

$1220=16 m+c$

So, our equations are

$980=14 m+c$

$1220=16 m+c$

From (A)

$980=14 m+c$

$980-14 m=c$

Putting value of $\mathrm{c}$ in $(\mathrm{B})$

$1220=16 m+980-14 m$

$1220-980=16 m-14 m$

$240=2 m$

$\dfrac{240}{2}=2$

$120=m$

$m=120$

Putting $m=120$ in $(A)$

$980=14 m+c$

$980-14 m=c$

$980-14(120)=c$

$980-1680=c$

$-700=c$

$c=-700$

Putting value of $m \& c$ in (1)

$D=m P+c$

D=120 P-700

Hence, the required equation is $\mathrm{D}=120 \mathrm{P}-700$

We need to find how many litres could he sell weekly at Rs $17 /$ litre i.e. we need to find $D$ when $P=17$

Putting $P=17$ in the equation

D=120 P-700

D=120(17)-700

D=2040-700

D=1340

Hence when price is Rs $17 /$ litre, 1340 litres of milk could be sold.

10. The line through the points $({\text{h}},3)$ and $(4,1)$ intersects the line $7x - 9y - 19 = 0$ a right angle. Find the value of $h$.

Ans: Given a line joining $({\text{h}},3)$ and $(4,1)$

Let $\left( {{x_1},\,{y_1}} \right){\text{ and }}\left( {{x_2},\,{y_2}} \right)$ be $({\text{h}},3)$ and $(4,1)$ respectively.

Find slope by two-point form,

$ {m_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} $

$ \Rightarrow \dfrac{{1 - 3}}{{4 - h}} = \dfrac{{ - 2}}{{4 - h}}$

Given line is $7x - 9y - 19 = 0$

${m_2} = \dfrac{{ - 7}}{{ - 9}}$

ATQ,

The lines intersect at right angles, i.e., product of slope $ = - 1$

$\,\,\therefore \,\,\,\,\,\,\,\,\,\,m{{\kern 1pt} _1} \times {m_2} = - 1$

$\left( {\dfrac{{ - 2}}{{4 - h}}} \right) \times \left( {\dfrac{7}{9}} \right) = - 1$

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,h = \dfrac{{22}}{9}$

11. Find the equations of the lines, which cut off intercepts on the axes whose sum and products are 1 and $ - 6$ respectively.

Ans: Let the intercepts made by the line on the axes be,

($x - $intercept) $ = a$

($y - $intercept) $ = b$

ATQ,

Sum of intercepts, i.e., $a + b = 1\,\,\,\,\,\,\,\,.....(i)$

Product of intercepts, i,e., $ab = - 6\,\,\,\,\,\,\,\,.....\left( {ii} \right)$

$b = 1 - a\quad [$ from $(i)]$

Put b in equation (ii)

$a(1 - a) = - 6$

$a - {a^2} = - 6$

${a^2} - a - 6 = 0$

$(a - 3)(a + 2) = 0$

$a = 3, - 2$

When $a = 3$

$b = - 2$

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

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

$2x - 3y - 6 = 0$

When $a = - 2$

$b = 3$

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

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

$3x - 2y + 6 = 0$

12. The slope of a line is double of the slope of another line. If the tangent of the angle between them is $\dfrac{1}{3}$, find the slopes of the lines.

Ans: Given,

The slope of a line is double of the slope of another line.

Tangent of the angle between them, i.e., $\tan \theta = \dfrac{1}{3}$

Let the slope of one line is $m$ and other line is 2m

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

$\dfrac{1}{3} = \left| {\dfrac{m}{{1 + 2{m^2}}}} \right|$

$ \pm \dfrac{1}{3} = \dfrac{m}{{1 + 2{m^2}}}$

$\dfrac{1}{3} = \dfrac{m}{{1 + 2{m^2}}}$

$2{m^2} - 3m + 1 = 0$

$2{m^2} - 2m - m + 1 = 0$

$2m(m - 1) - 1(m - 1) = 0$

$(m - 1)(2m - 1) = 0$

$m = 1,\,\,m = \dfrac{1}{2}$

$\dfrac{{ - 1}}{3} = \dfrac{m}{{1 + 2{m^2}}}$

$ - 1 - 2{m^2} = 3m$

$2{m^2} + 3m + 1 = 0$

$2{m^2} + 2m + m + 1 = 0$

$2m(m + 1) + 1(m + 1) = 0$

$(m + 1)(2m + 1) = 0$

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

13. Point $R(h,k)$ divides a line segment between the axes in the ratio $1:2$. Find the equation of the line.

Ans: It is given that $R(h,k)$ divides ${\text{AB}}$ in the ratio $1:2$

Required equation is $\dfrac{x}{a} + \dfrac{y}{b} = 1\,\,\,\,\,.....\left( i \right)$

$\therefore \left( {h,\,k} \right) = \left( {\dfrac{{2a}}{3},\,\dfrac{b}{3}} \right) $

$h = \dfrac{{2a}}{3}\,\,\,\,{\text{and }}\,\,\,k = \dfrac{b}{3} $

$\dfrac{{3h}}{2} = a\,\,\,{\text{and}}\,\,\,3k = b$

Put $a$ and $b$ in equation ……(i)

$\dfrac{x}{{\dfrac{{3h}}{2}}} + \dfrac{y}{{3k}} = 1 $

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

14. The Fahrenheit temperature F and absolute temperature K satisfy a linear equation. Given that ${\text{K}} = 273$ when ${\text{F}} = 32$ and that ${\text{K}} = 373$ when ${\text{F}} = 212$. Express ${\text{K}}$ in terms of ${\text{F}}$ and find the value of ${\text{F}}$ when ${\text{K}} = 0$

Ans: Let Fahrenheit temperature ${\text{F}}$ along $x$-axis and absolute temperature ${\text{K}}$ along $y$-axis.

Let $\left( {{x_1},\,{y_1}} \right){\text{ and }}\left( {{x_2},\,{y_2}} \right)$ be $\left( {273,\,32} \right)$ and $\left( {373,\,212} \right)$ respectively.

Using equation of straight line by two-point form,

$y - {y_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)$

$K - 273 = \dfrac{{373 - 273}}{{212 - 32}}(F - 32)$

$K - 273 = \dfrac{{10}}{{18}}(F - 32)$

When $K = 0$, then

$\,\,\,\,K = \dfrac{5}{9}(F - 32) + 273 $

$\therefore F = \dfrac{9}{5}\left( {K - 273} \right) + 32\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\,K = 0} \right] $

$\,\,\,\,\,\,\,\,\, = \dfrac{9}{5} \times \left( { - 273} \right) + 32 $

$\,\,\,\,\,\,\,\,\, = - 491.4 + 32 $

$\,\,\,\,\,\,\,\,\, = - 459.4$

15. If three points $(h,0),\,\,(a,b)$ and $(0,k)$ lie on a line, show that $\dfrac{a}{h} + \dfrac{b}{k} = 1$

Ans: Let $A(h,0),\,\,B(a,b)$ and ${\text{C}}(0,k)$

Since, the three points lie on the same line. Therefore,

Slope of ${\text{AB}} = $ slope of ${\text{BC}}$

$\Rightarrow \dfrac{b}{a-h}=\dfrac{k-b}{-a} $

$\Rightarrow (b)(-a)=(k-b)(a-h) $

$\Rightarrow(-a)(b)=ka-kh-ab+bh $

$\Rightarrow ka+bh=k h$

$\dfrac{{ak}}{{hk}} + \dfrac{{hb}}{{hk}} = 1$

$\dfrac{a}{h} + \dfrac{b}{k} = 1$

16. $P\left( {a,\,\,b} \right)$ is the midpoint of a line segment between axes. Show that the equation of the line is $\dfrac{x}{a} + \dfrac{y}{b} = 2$.

Ans: Given,

Mid-point of a line segment between axes $P\left( {a,\,\,b} \right)$.

ATQ,

Let P be the mid-point of $A\left( {c,\,0} \right)$ and $B\left( {0,\,d} \right)$

Required equation is $\dfrac{x}{a} + \dfrac{y}{b} = 1$ i.e., $\dfrac{x}{c} + \dfrac{y}{d} = 1$.

Where,

$x - $intercept = a

$y - $intercept = b

Coordinate of $P = \left( {\dfrac{c}{2},\dfrac{d}{2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because {\text{ mid - point}} = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)} \right]$

$\therefore \,\,(a,b) = \left( {\dfrac{c}{2},\dfrac{d}{2}} \right)$

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

$c = 2a$

$\dfrac{b}{1} = \dfrac{d}{2}$

$d = 2b$

Put the value of $c$ and $d$ in equation (i)

$\dfrac{x}{{2a}} + \dfrac{y}{{2b}} = 1$

$\,\,\,\,\,\dfrac{x}{a} + \dfrac{y}{b} = 2$

17. The line $ \bot $ to the line segment joining the points $(1,0)$ and $(2,3)$ divides it in the ratio $1:n$ find the equation of the line.

Ans: Given,

Coordinates of line segments.

Let $A\left( {1,\,\,0} \right)$ and $B\left( {2,\,\,3} \right)$. Ratio = $1:n$

Coordinate of $C\left( {\dfrac{{2 + n}}{{1 + n}} = \dfrac{3}{{1 + n}}} \right)$

Slope of AB,

${m_{AB}} = 3$

Slope of PQ,

${m_{PQ}} = - \dfrac{1}{3}$

Equation of ${\text{PQ}}$using one-point form,

$\left( {{x_1},\,{y_1}} \right) = \left( {\dfrac{{2 + n}}{{1 + n}} = \dfrac{3}{{1 + n}}} \right)$

$\,\,\,\,\,\,\,{m_{PQ}} = - \dfrac{1}{3}$

$y - {y_1} = m\left( {x - {x_1}} \right)$

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

$(n + 1)x + 3(n + 1)y - (n + 11) = 0$

Long Answer Question (6 Marks)

1. Find the values of $k$ for the line $(k - 3)x - \left( {4 - {k^2}} \right)y + {k^2} - 7k + 6 = 0$

(a). Parallel to the $x$-axis

(b). Parallel to $y$-axis

(c). Passing through the origin

Ans:

Given, equation of line $(k - 3)x - \left( {4 - {k^2}} \right)y + {k^2} - 7k + 6 = 0$

(a) The line is parallel to $x$-axis, if coefficient of $x = 0$

coefficient of $x = k - 3$

$\therefore \,\,k - 3 = 0$

$\,\,\,\,\,\,\,\,\,\,\,k = 3$

(b) The line is parallel to $y$-axis, if coefficient of $y = 0$

coefficient of $y = 4 - {k^2}$

$\therefore \,\,4 - {k^2} = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,k = \pm 2$

(c) line passes through the origin if $(0,0)$ lies on given equation,

$(k - 3) \cdot (0) - \left( {4 - {k^2}} \right)(0) + {k^2} - 7k + 6 = 0$

$(k - 6)(k - 1) = 0$

$k = 6,1$

2. If $p$ and $q$ are the lengths of $ \bot $ from the origin to the lines $x\cos \theta - y\sin \theta = k\cos 2\theta $ and $x\sec \theta + y\operatorname{cosec} \theta = k$ respectively, prove that ${p^2} + 4{q^2} = {k^2}$.

Ans:

Equation of lines are,

$x \cos \theta-y \sin \theta=k \cos 2 \theta \ldots . .(1)$

$\mathrm{x} \sec \theta+\mathrm{y} \csc \theta=\mathrm{k} \ldots .(2)$

The general equation is of the from ${\text{Ax}} + {\text{By}} + {\text{C}} = 0$.

The perpendicular distance $d$ of a line from a point $\left( {{{\text{x}}_1},{{\text{y}}_1}} \right)$ is given by,

${\text{d}} = $ $\dfrac{{\left| {{\text{A}}{{\text{x}}_1} + {\text{B}}{{\text{y}}_1} + {\text{C}}} \right|}}{{\sqrt {{{\text{A}}^2} + {{\text{B}}^2}} }}$

On comparing equation (1) to the general equation of line i.e.,

$Ax + By + C = 0$, we obtain

$A = \cos \theta ,\,\,{\text{B}} = - \sin \theta $, and ${\text{C}} = - {\text{k}}\cos 2\theta $

It is given that ${\text{p}}$ is the length of the perpendicular from $\left( {0,\,\,0} \right)$ to line (1)

$\therefore {\text{p}} = \dfrac{{|{\text{A}}(0) + {\text{B}}(0) + {\text{C}}|}}{{\sqrt {{{\text{A}}^2} + {{\text{B}}^2}} }}$

$\Rightarrow \dfrac{{|{\text{C}}|}}{{\sqrt {{{\text{A}}^2} + {{\text{B}}^2}} }} = \dfrac{{| - {\text{k}}\cos 2\theta |}}{{\sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } }}$

$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,| - {\text{k}}\cos 2\theta |\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \ldots ..(3)$

On comparing equation(2) to the general equation of line i.e.,

${\text{Ax}} + {\text{By}} + {\text{C}} = 0$, we obtain

$A = \sec \theta ,B = {\text{cosec}}\theta $, and $C = - {\text{k}}$

It is given that ${\text{q}}$ is the length of the perpendicular from $(0,0)$ to line (2) $\therefore {\text{q}} = \dfrac{{|{\text{A}}(0) + {\text{B}}(0) + {\text{C}}|}}{{\sqrt {{{\text{A}}^2} + {{\text{B}}^2}} }}$

$\Rightarrow \dfrac{{|{\text{C}}|}}{{\sqrt {{{\text{A}}^2} + {{\text{B}}^2}} }} = \dfrac{{| - {\text{k}}|}}{{\sqrt {{{\sec }^2}\theta + {{\csc }^2}\theta } }}\,\,\,\,\, \ldots ..(4)$

From (3) and (4) we have,

${{\text{p}}^2} + 4{{\text{q}}^2} = {\left( {| - {\text{k}}\cos 2\theta |} \right)^2} + 4{\left\{ {\dfrac{{| - {\text{k}}|}}{{\sqrt {{{\sec }^2}\theta + {{\csc }^2}\theta } }}} \right\}^2}$

$ = {{\text{k}}^2}{\cos ^2}2\theta + \dfrac{{4{{\text{k}}^2}}}{{\left( {{{\sec }^2}\theta + {{\csc }^2}\theta } \right)}}$

$ = {{\text{k}}^2}{\cos ^2}2\theta + \dfrac{{4{{\text{k}}^2}}}{{\left\{ {\dfrac{1}{{{{\cos }^2}\theta }} + \dfrac{1}{{{{\sin }^2}\theta }}} \right\}}}$

$ = {{\text{k}}^2}{\cos ^2}2\theta + \dfrac{{4{{\text{k}}^2}}}{{\left\{ {\dfrac{{{{\sin }^2} + {{\cos }^2}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }}} \right\}}}$

$ = {{\text{k}}^2}{\cos ^2}2\theta + 4{{\text{k}}^2}{\sin ^2}\theta {\cos ^2}\theta $

$ = {{\text{k}}^2}{\cos ^2}2\theta + {{\text{k}}^2}{(2\sin \theta \cos \theta )^2}$

$ = {{\text{k}}^2}{\cos ^2}2\theta + {{\text{k}}^2}{\sin ^2}2\theta $

$ = {{\text{k}}^2}\left( {{{\cos }^2}2\theta + {{\sin }^2}2\theta } \right) = {{\text{k}}^2}\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \,\,{{\cos }^2}\theta + {{\sin }^2}\theta = 1} \right]$

Hence proved ${{\text{p}}^2} + 4{{\text{q}}^2} = {{\text{k}}^2}$.

3. Prove that the product of the $ \bot $ drawn from the points $\left( {\sqrt {{a^2} - {b^2}} ,0} \right)$ and $\left( { - \sqrt {{a^2} - {b^2}} ,0} \right)$ to the line $\dfrac{x}{a}\cos \theta + \dfrac{y}{b}\sin \theta = 1$ is ${b^2}$.

Ans: Given equation of line is $\dfrac{x}{a}\cos \theta + \dfrac{y}{b}\sin \theta = 1$

Let ${p_1}$ be the distance from $\left( {\sqrt {{a^2} - {b^2}} ,0} \right)$ to the given line,

${p_1} = \dfrac{{\mid \dfrac{{\sqrt {{a^2} - {b^2}} }}{a} \cdot \cos \theta - 1}}{{\sqrt {{{\left( {\dfrac{{\cos \theta }}{a}} \right)}^2} + {{\left( {\dfrac{{\sin \theta }}{b}} \right)}^2}} }}\,\,\,\,\,\,\left[ {\because \bot {\text{ from the points }}\sqrt {{a^2} - {b^2}} ,0} \right]$

Similarly,

${p_2}$ be the distance from $\left( { - \sqrt {{a^2} - {b^2}} ,0} \right)$ to given line,

${p_2} = \dfrac{{\left| { - \dfrac{{\sqrt {{a^2} - {b^2}} }}{a}\cos \theta - 1} \right|}}{{\sqrt {{{\left( {\dfrac{{\cos \theta }}{a}} \right)}^2} + {{\left( {\dfrac{{\sin \theta }}{b}} \right)}^2}} }}$

Product of perpendicular lines, i.e., ${p_1}.{p_2}$

${p_1}{p_2} = \dfrac{{\left| {\left( {\dfrac{{\sqrt {{a^2} - {b^2}} }}{a}\cos \theta - 1} \right)\left( { - \dfrac{{\sqrt {{a^2} - {b^2}} }}{a}\cos \theta - 1} \right)} \right|}}{{\dfrac{{{{\cos }^2}\theta }}{{{a^2}}} + \dfrac{{{{\sin }^2}\theta }}{{{b^2}}}}}$

$ = \dfrac{{\left| {\left( {\dfrac{{{a^2} - {b^2}}}{{{a^2}}}} \right) \cdot {{\cos }^2}\theta - 1} \right|}}{{\dfrac{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}{{{a^2}{b^2}}}}}$

$ = \dfrac{{\left| {{a^2}{{\cos }^2}\theta - {b^2}{{\cos }^2}\theta - {a^2}} \right|{a^2}{b^2}}}{{{a^2}\left( {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } \right)}}$

$= \dfrac{{\left| { - \left( {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } \right)} \right|{b^2}}}{{{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta }}\quad \left[ {\because \,\,{a^2}{{\cos }^2}\theta - {a^2} = {a^2}\left( {{{\cos }^2}\theta - 1} \right)} \right]$

$= \dfrac{{\left( {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } \right){b^2}}}{{{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta }}$

$= {b^2}$

Hence proved that the product of the $ \bot $ drawn from the points $\left( {\sqrt {{a^2} - {b^2}} ,0} \right)$ and $\left( { - \sqrt {{a^2} - {b^2}} ,0} \right)$ to the line $\dfrac{x}{a}\cos \theta + \dfrac{y}{b}\sin \theta = 1$ is ${b^2}$.

4. Find equation of the line mid way between the parallel lines $9x + 6y - 7 = 0$ and $3x + 2y + 6 = 0$

Ans: The given equations of parallel lines are

$9x + 6y - 7 = 0$

$3\left( {3x + 2y - \dfrac{7}{3}} \right) = 0$

$3x + 2y - \dfrac{7}{3} = 0 \ldots \ldots (i)$

$3x + 2y + 6 = 0 \ldots ..(ii)$

Let the eq. of the line mid way between the parallel lines (i) and (ii) be

$3x + 2y + k = 0 \ldots \ldots .(iii)$

ATQ,

Distance between (i) and (iii) = distance between (ii) and (iii)

Distance between two parallel line,

$d = \dfrac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {{a^2} + {b^2}} }}$

$\left| {\dfrac{{k + \dfrac{7}{3}}}{{\sqrt {9 + 4} }}} \right| = \left| {\dfrac{{k - 6}}{{\sqrt {9 + 4} }}} \right|\left[ {\because d = \dfrac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {{a^2} + {b^2}} }}} \right]$

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

$k = \dfrac{{11}}{6}$

Substituting the value of $k$ in equation (iii)

Therefore, the required equation is $3x + 2y + \dfrac{{11}}{6} = 0$.

5. Assuming that straight lines work as the plane mirror for a point, find the image of the point $(1,2)$ in the line $x - 3y + 4 = 0$

Ans: Let $Q(h,k)$ is the image of the point $P(1,2)$ in the line $x - 3y + 4 = 0 \ldots \ldots (i)$

Coordinate of midpoint of $PQ = \left( {\dfrac{{h + 1}}{2},\dfrac{{k + 2}}{2}} \right)$

This point will satisfy the eq. .......(i)

$\left( {\dfrac{{h + 1}}{2}} \right) - 3\left( {\dfrac{{k + 2}}{2}} \right) + 4 = 0$

$\dfrac{{h + 1}}{2} - \dfrac{{3k + 6}}{2} + 4 = 0$

$h - 3k = - 3 \ldots \ldots .(ii)$

Since, the object and the line are perpendicular. Therefore,

(Slope of line ${\text{PQ}}) \times $ (slope of line $x - 3y + 4 = 0) = - 1$

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

$3h + k = 5 \ldots \ldots .(iii)$

On solving (ii) and (iii)

$h = \dfrac{6}{5}$ and $k = \dfrac{7}{5}$

6. A person standing at the junction (crossing) of two straight paths represented by the equations $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$ wants to reach the path whose equation is $6x - 7y + 8 = 0$ in the least time. Find the equation of the path that he should follow.

Ans: The given equations of parallel lines are,

$2x - 3y - 4 = 0$....... (i)

$3x + 4y - 5 = 0 \ldots \ldots (ii)$

Person wants to reach the path whose equation is,

$6x - 7y + 8 = 0 \ldots \ldots (iii)$

On solving eq. (i) and (ii)

we get $\left( {\dfrac{{31}}{{17}},\dfrac{{ - 2}}{{17}}} \right)$

To reach the line (iii) in least time the man must move along the perpendicular from crossing point $\left( {\dfrac{{31}}{{17}},\dfrac{{ - 2}}{{17}}} \right)$ to (iii) line,

Slope of line(iii) is $\dfrac{6}{7}$

Therefore, the slope of the required path,

Slope of the required path $ \times $Slope of line(iii) $ = - 1$

Slope of the required path $ \times $$\dfrac{6}{7}$$ = - 1$

Slope of the required path$ = - \dfrac{7}{6}$

Therefore, the equation of the lie using one-point form,

$y - {y_1} = m\left( {x - {x_1}} \right)$

$\,\,\,\,y - \left( { - \dfrac{2}{{17}}} \right) = - \dfrac{7}{6}\left( {x - \dfrac{{31}}{{17}}} \right)$

$119x + 102y = 205$

7. A line is such that its segment between the lines $5x - y + 4 = 0$ and $3x + 4y - 4 = 0$ is bisected at the point $(1,5)$ obtains its equation.

Ans: $P\left( {{x_1},\,\,{y_1}} \right)$ lies on $5x - y + 4 = 0$

$5{x_1} - {y_1} + 4 = 0$

and $Q\left( {{x_2}{y_2}} \right)$ lies on $3x + 4y - 4 = 0$

$3{x_2} + 4{y_2} - 4 = 0$

On simplifying,

${y_1} = 5{x_1} + 4$

${y_2} = \dfrac{{4 - 3{x_2}}}{4}$

Since ${\text{R}}$ is the mid-point of PQ,

$\dfrac{{{x_1} + {x_2}}}{2} = 1,\dfrac{{{y_1} + {y_2}}}{2} = 5$

${x_1} + {x_2} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( i \right)$

${y_1} + {y_2} = 10\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( {ii} \right)$

On substituting the values of ${y_1}{\text{ and }}{y_2}$ from above in equation (ii),

$5{x_1} + 4 + \dfrac{{4 - 3{x_2}}}{4} = 10$

$20{x_1} + 16 + 4 - 3{x_2} = 40$

$20{x_1} - 3{x_2} = 20\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( {iii} \right)$

On solving equation (i) and (iii), we get

${x_1} = \dfrac{{26}}{{23}},{x_2} = \dfrac{{20}}{{23}}$

and ${y_1} = \dfrac{{222}}{{23}},{y_2} = \dfrac{8}{{23}}$

Hence,

$P\left( {\dfrac{{26}}{{23}},\,\,\dfrac{{222}}{{23}}} \right){\text{ and }}Q\left( {\dfrac{{20}}{{23}},\,\dfrac{8}{{23}}} \right)$

Equation of line PQ using two-point form,

$y - {y_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)$

$y - \dfrac{{222}}{{23}} = \dfrac{{\dfrac{8}{{23}} - \dfrac{{222}}{{23}}}}{{\dfrac{{20}}{{23}} - \dfrac{{26}}{{23}}}}\left( {x - \dfrac{{26}}{{23}}} \right)$

$107x - 3y - 92 = 0$

8. Find the equations of the lines which pass through the point $(4,5)$ and make equal angles with the lines $5x - 12y + 6 = 0$ and $3x - 4y - 7 = 0$

Ans: Let the equation of line passing through the point $(4,5)$ be,

$y - 5 = m\left( {x - 4} \right)\,\,\,\,\,.....\left( i \right)$

Given an equation of two lines $5x - 12y + 6 = 0$ and $3x - 4y - 7 = 0$.

slope of line $5x - 12y + 6 = 0$ is ${m_1} = \dfrac{5}{{12}}$

slope of line $3x - 4y - 7 = 0$ is ${m_2} = \dfrac{3}{4}$

Since, the line makes equal angles. Therefore,

$\tan \theta = \dfrac{{m - {m_1}}}{{1 + m.{m_1}}}{\text{ and }}\tan \theta = \dfrac{{m - {m_2}}}{{1 - m.{m_2}}}$

$\dfrac{{m - \dfrac{5}{{12}}}}{{1 + m.\dfrac{5}{{12}}}} = \dfrac{{m - \dfrac{3}{4}}}{{1 + m.\dfrac{3}{4}}}$

On solving equation (i) and (iii), we get

$m = \dfrac{{ - 7}}{4}\;\;{\text{and }}m = \dfrac{4}{7}$

Putting the above values in equation (i),

Required equations are

$y - 5 = \dfrac{4}{7}(x - 4){\text{ and }}y - 5 = \dfrac{{ - 7}}{4}(x - 4)$

$4x - 7y + 19 = 0\,\,\,\,{\text{and}}\,\,\,7x + 4y - 48 = 0$

## Important Questions from Straight Lines (Short, Long & Practice)

### Short Answer Type Questions

1. Find the slope of the lines passing through the point (3,-2) and (-1,4)

2. Write the equation of the line through the points (1,-1) and (3,5)

3. Find the measure of the angle between the lines x+y+7=0 and x-y+1=0

### Long Answer Type Questions

1. Find the equation to the straight line which passes through the point (3,4) and has an intercept on the axes equal in magnitude but opposite in sign.

2. Find the equations of the lines, which cut off intercepts on the axes whose sum and product are 1 and -6 respectively.

3. Find equation of the line passing through the point (2,2) and cutting off intercepts on the axes whose sum is 9

### Practice Questions

1. Find the equation of a straight line parallel to y-axis and passing through the point (4,-2)

2. Find the slope of the line, which makes an angle of 30o with the positive direction of y-axis measured anticlockwise.

3. Find the equation of the line intersecting the x-axis at a distance of 3 units to the left of origin with slope -2.

## Key Features of Important Questions for Class 11 Maths Chapter 10 - Straight Lines

All the questions are curated as per examination point of view to help students score better.

Solutions are explained in a step by step manner for all questions.

All solutions are easy to understand and learn as they are clearly written by subject experts to match the curriculum.

These important questions help in developing a good conceptual foundation for students, which is important in the final stages of preparation for board and competitive exams.

These solutions are absolutely free and available in a PDF format.

## Important Related Links for CBSE Class 11

CBSE Class 11 Study Materials |

## Conclusion

Vedantu's **Important Questions for Class 11 Maths Chapter 10** - Straight Lines prove to be an invaluable resource for students seeking comprehensive preparation. By meticulously curating essential concepts and problem-solving techniques, Vedantu empowers learners to master this fundamental topic with confidence. The well-structured questions encourage critical thinking and enhance problem-solving skills, enabling students to grasp the intricacies of straight lines effortlessly. Additionally, the platform's interactive approach fosters a deeper understanding of mathematical principles, promoting a strong foundation for future studies. Vedantu's commitment to academic excellence and accessibility makes it an ideal companion for students aspiring to excel in mathematics. With Vedantu's guidance, navigating the complexities of straight lines becomes an enriching and rewarding experience.

## FAQs on Important Questions for Class 11 Maths Chapter 10 - Straight Lines

1. What is the significance of Vedantu’s extra questions for Class 11 Maths Chapter 10 Straight Lines?

Vedantu’s extra questions for CBSE Class 11 Maths Chapter 10 Straight Lines are helpful during the exam preparation. These important questions from Class 11 Maths Chapter 10 allow students to read the chapter thoroughly. Working on these questions will make students familiar with all types of questions that are important from an examination point of view. This PDF covers all the important topics of the chapters. These questions are based on the exam pattern and are added after referring to previous year question papers. The free PDF of CBSE Maths Class 11 Chapter 10 can be used at the time of revision and they are extremely helpful in scoring well in the paper.

2. What are the important topics of Class 11 Maths Chapter 10 Straight Lines?

Important topics of Class 11 Maths Chapter 10 Straight Lines:

Slope of a Line

Slope of a line when coordinates of any two points on the line are given

Conditions for parallelism and perpendicularity of lines in terms of their slopes

Angle between two lines

Collinearity of three points

Various Forms of the Equation of a Line

Horizontal and vertical lines

Point-slope form

Two-point form

Slope-intercept form

Intercept form

Normal form

General Equation of a Line

Different forms of Ax + By + C = 0

Distance of a Point From a Line

Distance between two parallel lines

3. Where can I find important questions for Class 11 Maths Chapter 10 Straight Lines?

Vedantu offers a well-prepared set of Important Questions for Class 11 Maths Chapter 10 Straight Lines as well as other chapters. Our teacher selects questions based on the exam pattern and questions from the previous papers. These solutions are also solved by subject matter experts to provide a thorough understanding of the chapter. These are definitely helpful in exam preparation and revisions during exams.

4. What are some of the important properties of straight lines?

Some of the important properties of straight lines are:

Two lines are perpendicular if the angle between them is 90°.

Two lines are parallel if the angle between them is 0°.

The product of the slopes of two perpendicular lines is equal to -1.

What are some of the applications of straight lines?

Straight lines have many applications in the real world, such as:

Building roads and bridges

Designing machines

Creating maps

Constructing buildings

5. How can I prepare for the Class 11 Maths exam on Straight Lines?

Here are some tips on how to prepare for the Class 11 Maths exam on Straight Lines:

Make sure you understand the basic concepts of straight lines.

Practice solving problems on straight lines.

Refer to the NCERT textbook, other standard textbooks and Vedantu’s notes

Take online tests and mock exams provided by vedantu.

6. How can we find the angle between two straight lines?

The angle between two straight lines can be defined as the angle between their direction vectors. The direction vector of a line is a vector that points in the direction of the line. The angle between two lines can be found using the dot product of their direction vectors.