NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines (Ex 10.2) Exercise 10.2

VSAT 2022

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines (Ex 10.2) Exercise 10.2

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.2 (Ex 10.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 10 Straight Lines Exercise 10.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

Competitive Exams after 12th Science
Do you need help with your Homework? Are you preparing for Exams?
Study without Internet (Offline)

Download PDF of NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines (Ex 10.2) Exercise 10.2

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.2

Exercise 10.2

1. Write the equation for the x and y-axes.

Ans:The coordinate of axis Y for each point that is on x-axis will be 0.

Therefore, the equation of the x-axis is$y = 0$ .

The coordinate of axis X for each point that is on the y-axis will be 0. 

Therefore, the equation of the y-axis is$x = 0$ .


2. Find the equation of the line which passes through the point $\left( { - 4,3} \right)$ with slope${\scriptstyle{}^{1}/{}_{2}}$.

Ans:The equation of the line which pass through point$\left( {{x}_{0}},{{y}_{0}} \right)$, 

with slope is m, is- $\left( {y - {y_0}} \right) = m\left( {x - {x_0}} \right)$

Thus, the equation of the line which pass through point$\left( { - 4,3} \right)$, with slope is ${\scriptstyle{}^{1}/{}_{2}}$.

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

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

$\Rightarrow 2y-6=x+4$

$\therefore x-2y+10=0$


3. Find the equation of the line which passes through (0, 0) with slope m.

Ans:We know that the equation of the line passing through point$\left( {{x_0},{y_0}} \right)$, 

 Whose slope is m, is- $\left( {y - {y_0}} \right) = m\left( {x - {x_0}} \right)$

 Thus, the equation of the line passing through point$\left( 0,0 \right)$ , whose slope is m, is-

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

$\Rightarrow y=mx$ 

Hence, the equation of the line passing through origin is $y = mx$ .

        

4. Find the equation of the line which passes through $\left( {2,2\sqrt 3 } \right)$and is inclined with the X-axis at an angle of ${75^0}$ 

Ans:The slope of the line that inclines with the x-axis at an angle ${{75}^{0}}$is $m=\tan {{75}^{0}}$ 

$\Rightarrow m=\tan \left( {{45}^{0}}+{{30}^{0}} \right)$ 

$\Rightarrow m=\frac{\tan {{45}^{0}}+\tan {{30}^{0}}}{1-\tan {{45}^{0}}.\tan {{30}^{0}}}$

$\Rightarrow m=\frac{1+\frac{1}{\sqrt{3}}}{1-1.\frac{1}{\sqrt{3}}}$ 

$\Rightarrow m=\frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}}$

$\Rightarrow m=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

$\Rightarrow m=\frac{\sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$

$\Rightarrow m=\frac{{{\left( \sqrt{3}+1 \right)}^{2}}}{{{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}}}$

$\Rightarrow m=\frac{{{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}+2\sqrt{3}}{3-1}$

$\Rightarrow m=\frac{3+1+2\sqrt{3}}{2}$

$\Rightarrow m=\frac{4+2\sqrt{3}}{2}$

$\Rightarrow m=\frac{2(2+\sqrt{3})}{2}$

$\Rightarrow m=2+\sqrt{3}$

It is known that the equation of the line passing through point$\left( {{x}_{0}},{{y}_{0}} \right)$, 

Whose slope is m, is- $\left( y-{{y}_{0}} \right)=m\left( x-{{x}_{0}} \right)$

Thus, if a line passes through$\left( 2,2\sqrt{3} \right)$ and inclines with the x-axis at an angle of ${{75}^{0}}$ , then the equation of the line is given as-

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

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

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

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

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

$\left( 2+\sqrt{3} \right)x-y-4=0$

Therefore, the equation of the line is $\left( 2+\sqrt{3} \right)x-y-4=0$.


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

Ans:We know that-

 When a line makes x-intercept d with slope m, 

Then the equation of the line is given as-$Y = m\left( {x - d} \right)$ 

The line that intersects x-axis at a distance of 3 units to the left of the origin, $d =  - 3$ 

The slope of the line is will be $m=-2$ 

Hence, the required equation of the given line is-

$Y =  - 2\left[ {x - \left( { - 3} \right)} \right]$ 

$\Rightarrow Y=-2x-6$ 

$i.e.2x + y + 6 = 0$         


6. Find the equation of the line which intersects the y-axis at a distance of 2 units above the origin and makes an angle of ${{30}^{0}}$ with the positive direction of the x-axis.

Ans:If a line with slope m makes an intercept c on y-axis, then the equation of the line will be-$Y=mx+c$ 

Here, we have

 and $m = \tan {30^0}$

Therefore, $m={\scriptstyle{}^{1}/{}_{\sqrt{3}}}$

Hence, the required equation of the given line is-

$y={\scriptstyle{}^{1}/{}_{\sqrt{3}}}x+2$ 

$\Rightarrow y=\frac{x+2\sqrt{3}}{\sqrt{3}}$ 

$\Rightarrow \sqrt{3}y=x+2\sqrt{3}$ 

$\Rightarrow x-2\sqrt{3}+2\sqrt{3}=0$ 


7. Find the slope of the line, which makes an angle of ${30^0}$  with the positive direction of y-axis measured anticlockwise.

Ans:It is known that the equation of the line passes through points $\left( {{x_1},{y_1}} \right)$and $\left( {{x_2},{y_2}} \right)$is-

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

Therefore, the equation of the line passing through the points $\left( { - 1,1} \right)$  and $\left( {2, - 4} \right)$is-

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

$\Rightarrow \left( {y - 1} \right) = \frac{{ - 5}}{3}\left( {x + 1} \right)$ 

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

$\Rightarrow 3y - 3 =  - 5x - 5$ 

$\Rightarrow 5x + 3y + 2 = 0$ 


8. Find the equation of the line which is at a perpendicular distance of 5 units from the origin and the angle made by the perpendicular with the positive x-axis is${30^0}$.

Ans:If the length of the normal from the origin to a line is p and \[\omega \] is the angle made by the normal with the positive direction of the x-axis, then the equation of the line given by 

$x\cos \omega +y\sin \omega =p$ 

Here,$p=5$units and $\omega ={{30}^{0}}$ 

Thus, the required equation of the given line is-

$x\cos {{30}^{0}}+y\sin {{30}^{0}}=5$ 

\[x\left( \frac{\sqrt{3}}{2} \right)+y\left( \frac{1}{2} \right)=5\]

\[x\left( \sqrt{3} \right)+y(1)=5\times 2\]

\[\sqrt{3}x+y=10\]

\[\sqrt{3}x+y-10=0\]

Hence, the equation of the line is\[\sqrt{3}x+y-10=0\]. 


9.  The vertices of$\Delta PQR$are $P\left( {2,1} \right),Q\left( { - 2,3} \right)$ and$R\left( 4,5 \right)$. Find the equation of the median through the vertex R.

Ans:It is given that the vertices of$\Delta PQR$are$P\left( 2,1 \right),Q\left( -2,3 \right)$ and $R\left( {4,5} \right)$Let RL be the median through vertex R.

Accordingly, L is the mid-point of PQ.

From mid-point formula, the coordinates of point L will be-

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

It is known that the equation of the line passing through a point.

$\left( {{x}_{1}},{{y}_{1}} \right)=\left( 4,5 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( 0,2 \right)$ 

Hence, we can put values as-

$y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$ 

$\Rightarrow \left( y-5 \right)=\frac{2-5}{0-4}\left( x-4 \right)$ 

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

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

$\Rightarrow 3x - 4y + 8 = 0$ 

Hence, the required equation of the median through vertex R is$3x-4y+8=0$.

              

10.  Find the equation of the line passing through $\left( -3,5 \right)$ and perpendicular to the line through the points $\left( 2,5 \right)$ and $\left( { - 3,6} \right)$ 

Ans:The slope of the line joining the points $\left( 2,5 \right)$ and $\left( { - 3,6} \right)$is 

$m = \frac{{6 - 5}}{{ - 3 - 2}}$ 

$m = \frac{1}{{ - 5}}$ 

Two non-vertical lines will be perpendicular to each other if and only if their slopes

are negative reciprocals of each other.

Therefore, slope of the line perpendicular to the line through the points $\left( 2,5 \right)$ and $\left( { - 3,6} \right)$is equal to

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

Now, the equation of the line passing through point$\left( { - 3,5} \right)$, whose slope is 5, is-

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

$\Rightarrow \left( {y - 5} \right) = 5\left( {x + 3} \right)$ 

$\Rightarrow y-5=5x+15$ 

$\Rightarrow 5x-y+20=0$ 

Hence, the required equation of line is-$5x-y+20=0$

        

11. A line perpendicular to the line segment joining the points $\left( 1,0 \right)$ and $\left( {2,3} \right)$ divides it in the ratio 1 : n. Find the equation of the line.

Ans:According to the section formula, the coordinates of the point that divides the line segment joining the points $\left( {1,0} \right)$ and $\left( 2,3 \right)$in the ratio 1 : n is given by

 $\left( \frac{n\left( 1 \right)+1\left( 2 \right)}{1+n},\frac{n\left( 0 \right)+1\left( 3 \right)}{1+n} \right)=\left( \frac{n+2}{n+1},\frac{3}{n+1} \right)$                                                                                           

The slope of the line joining the points $\left( 1,0 \right)$ and $\left( {2,3} \right)$is-

$m = \frac{{3 - 0}}{{2 - 1}} = 3$ 

We know that two non-vertical lines will be perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line that is perpendicular to the line joining the points$\left( {1,0} \right)$ and $\left( {2,3} \right)$is-

$\Rightarrow  - \frac{1}{m} =  - \frac{1}{3}$       

Hence,the equation of the line passing through   

$\left( {\frac{{n + 2}}{{n + 1}},\frac{3}{{n + 1}}} \right)$   and whose slope is $- \frac{1}{3}$is given by-

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

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

$\Rightarrow 3\left( n+1 \right)y-9=-\left( n+1 \right)x+n+2$ 

$\therefore \left( {1 + n} \right)x + 3\left( {1 + n} \right)y = n + 11$ 


12. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the points$\left( {2,3} \right)$.

Ans:The equation of a line in the intercept form is-

$\frac{x}{a} + \frac{y}{b} = 1\,\,\,\,......\left( i \right)$ 

Here, a and b will be intercepted on x and y axes respectively.

Then the line cuts off equal intercepts on both the axes. This means that$a = b$ .

Accordingly, equation (i) reduces to-

$\Rightarrow \frac{x}{a} + \frac{y}{b} = 1\,$

$\Rightarrow x + y = a\,\,......\left( {ii} \right)$ 

Since the given line passes through point$\left( 2,3 \right)$,equation (ii) reduces to 

$\Rightarrow 2 + 3 = a$ ,

i.e. $a = 5$ 

Now, we put the value of a in equation (ii), we obtain-

$\Rightarrow x+y=5$ 

Hence, the required equation of the line.      


13. Find the equation of the line passing through the points $\left( 2,2 \right)$ and cutting off intercepts on the axes whose sum is 9.

Ans:The equation of a line in the intercept form is-

$\frac{x}{a} + \frac{y}{b} = 1\,\,\,\,......\left( i \right)$

 Here, a and b will be intercepts on x and y axes respectively.    

 It is given that-

 $\Rightarrow a + b = 9$,

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

 Now, from equation (i) and (ii), we obtain-

$\frac{x}{a} + \frac{y}{{9 - a}} = 1\,\,\,\,......\left( {iii} \right)$

We have the line passes through point$\left( {2,2} \right)$.

Therefore equation (iii) reduces to-

$\Rightarrow \frac{2}{a}+\frac{2}{9-a}=1$ 

$\Rightarrow 2\left( {\frac{1}{a} + \frac{1}{{9 - a}}} \right) = 1$ 

$\Rightarrow 2\left( {\frac{{9 - a + a}}{{a\left( {9 - a} \right)}}} \right) = 1$ 

$\Rightarrow \frac{{18}}{{9a - {a^2}}} = 1$ 

$\Rightarrow 18=9a-{{a}^{2}}$

$\Rightarrow {{a}^{2}}-9a+18=0$ 

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

$\Rightarrow a\left( {a - 6} \right) - 3\left( {a - 6} \right) = 0$ 

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

$\Rightarrow a=6\,or\,a=3$ 

If $a = 6$ then $\Rightarrow b = 9 - 6 = 3$

Then equation of the line is- 

$\Rightarrow \frac{x}{6} + \frac{y}{3} = 1$ 

$\Rightarrow x + 2y - 6 = 0$ 

And if $a = 3$ then $\Rightarrow b=9-3=6$ 

Then equation of the line is-

$\Rightarrow \frac{x}{3} + \frac{y}{6} = 1$

$\Rightarrow 2x + y - 6 = 0$      


14. Find equation of the line through the points $\left( 0,2 \right)$  making an angle$\frac{2\pi }{3}$  with the positive X-axis. Also, find the equation of the line parallel to it and crossing the y-axis at a distance of 2units below the origin.

Ans:The slope of the line making an angle\[\frac{2\pi }{3}\] with the positive x-axis is-

$m = \tan \left( {\frac{{2\pi }}{3}} \right) =  - \sqrt 3$ 

 Now, the equation of the line passing through points$\left( {0,2} \right)$ and having a slope is $- \sqrt 3$

$\left( y-2 \right)=-\sqrt{3}\left( x-0 \right)$ 

i.e. $\sqrt{3}x+y-2=0$ 

The slope of line parallel to the line-$\sqrt 3 x + y - 2 = 0$is $- \sqrt 3$

We have the line parallel to line$\sqrt 3 x + y - 2 = 0$crosses the y-axis 2 units below the origin.

i.e. it passes through point $\left( {0,2} \right)$

Hence, the equation of line passing through the point $\left( {0,2} \right)$and having a slope $- \sqrt 3$ is-

$\Rightarrow y-\left( -2 \right)=-\sqrt{3}\left( x-0 \right)$ 

$\Rightarrow y + 2 =  - \sqrt 3 x$ 

$\therefore \sqrt{3}x+y+2=0$


15. The perpendicular from the origin to a line meets it at the point $\left( -2,9 \right)$, find the equation of the line.

Ans:The slope of the line joining the origin$\left( {0,0} \right)$and point $\left( { - 2,9} \right)$is -

${{m}_{1}}=\frac{9-0}{-2-0}=\frac{-9}{2}$ 

Accordingly, the slope of the line perpendicular to the line joining the origin and points$\left( { - 2,9} \right)$ is-

${m_2} =  - \frac{1}{{{m_1}}}$ 

$\Rightarrow {{m}_{2}}=-\frac{1}{\left( -\frac{9}{2} \right)}=\frac{2}{9}$ 

Now, the equation of the line passing through point $\left( { - 2,9} \right)$ and having a slope ${m_2}$ is-

$\Rightarrow \left( {y - 9} \right) = \frac{2}{9}\left( {x + 2} \right)$ 

$\Rightarrow 9y - 81 = 2x + 4$ 

$\therefore 2x - 9y + 85 = 0$ 


16. The length L (in centimeter) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if $L = 124.942$when$C = 20$ and $L = 125.134$ when$C=110$ , express L in terms of C.

Ans:It is given that when$C = 20$, the value of L is 124.942, whereas when$C = 110$, the value of L is 125.134.

Accordingly, points $\left( {20,124.942} \right)$ and $\left( {110,125.134} \right)$satisfy the linear relation between L and C.

If we assume C along the x-axis and L along the y-axis, we get two points 

i.e., $\left( {20,124.942} \right)$and$\left( {110,125.134} \right)$in the XY plane.

Therefore, the linear relation between L and C is the equation of the line passing through the points $\left( {20,124.942} \right)$and$\left( {110,125.134} \right)$

$\Rightarrow \left( {L - 124.942} \right) = \frac{{125.134 - 124.942}}{{110 - 20}}\left( {C - 20} \right)$                                  $\Rightarrow \left( {L - 124.942} \right) = \frac{{0.192}}{{90}}\left( {C - 20} \right)$

i.e. $L = \frac{{0.192}}{{90}}\left( {C - 20} \right) + 124.942$

This is required linear relation. 

 

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

Ans:We have relation between selling price and demand is linear.

Assuming selling price per liter along the x-axis and demand along the y-axis, we have two points i.e.,$\left( {14.980} \right)$and$\left( 16.1220 \right)$in the XY plane that satisfy the linear relationship between selling price and demand.

Therefore, the line passing through points$\left( {14.980} \right)$ and$\left( {16.1220} \right)$is-

$\Rightarrow \left( {y - 980} \right) = \frac{{1220 - 980}}{{16 - 14}}\left( {x - 14} \right)$

$\Rightarrow \left( {y - 980} \right) = \frac{{240}}{2}\left( {x - 14} \right)$

$\Rightarrow \left( {y - 980} \right) = 120\left( {x - 14} \right)$

         \[\therefore y = 120\left( {x - 14} \right) + 980\] 

When $x = 17/liter$ 

Therefore, $y = 120\left( {17 - 14} \right) + 980$ 

$\Rightarrow y = 120x3 + 980$ 

$\Rightarrow y = 360 + 980$ 

$\therefore y=1340$ 

Thus, the owner of the milk store could sell 1340 liters of milk weekly at Rs$17/liter


18. P (a, b) is the mid-point of a line segment between axes. Show that the equation of the line is$\frac{x}{a}+\frac{y}{b}=2$ 

Ans:Let AB be the line segment between the axes and let $P\left( {a,b} \right)$ be its mid-point

  Let the coordinates of A and B be $\left( 0,y \right)$ and $\left( x,0 \right)$ respectively.

  Since $P\left( a,b \right)$ is the mid- point of AB.

$\Rightarrow \left( \frac{0+x}{2},\frac{y+0}{2} \right)=\left( a,b \right)$

$\Rightarrow \left( \frac{x}{2},\frac{y}{2} \right)=\left( a,b \right)$ 

$\Rightarrow \frac{x}{2}=a$ and $\frac{y}{2} = b$ 

$\therefore x = 2a$ and $y=2b$ 

Thus, the respective coordinates of A and B are $\left( {0,2b} \right)$and $\left( 2a,0 \right)$. 

The equation of the line passing through points $\left( {0,2b} \right)$and$\left( {2a,0} \right)$is,

$\Rightarrow \left( {y - 2b} \right) = \frac{{\left( {0 - 2b} \right)}}{{\left( {2a - 0} \right)}}\left( {x - 0} \right)$ 

$\Rightarrow y - 2b = \frac{{ - 2b}}{{2a}}\left( x \right)$ 

$\Rightarrow a\left( {y - 2b} \right) =  - bx$ 

$\Rightarrow ay-2ab=-bx$ 

i.e., $bx+ay = 2ab$ 

We divide both sides by a b, and get-   $bx+ay = 2ab$

$\Rightarrow \frac{{bx}}{{ab}} + \frac{{ay}}{{ab}} = \frac{{2ab}}{{ab}}$ 

$\Rightarrow \frac{x}{a} + \frac{y}{b} = 2$ 


19.  Point $R\left( {h,k} \right)$ divides a line segment between the axes in the ratio$1:2$ . Find equation of the line.

Ans:Let AB be the line segment between the axes such that point$R\left( h,k \right)$ divides AB in the ratio 

 $1:2$.

  Let the respective coordinates of A and B be $\left( {x,0} \right)$ and $\left( {0,y} \right)$. 

  Since point $R\left( {h,k} \right)$divides AB in the ratio$1:2$, according to the section formula,

$\Rightarrow \left( {h,k} \right) = \left( {\frac{{1 \times 0 + 2 \times x}}{{1 + 2}},\frac{{1 \times y + 2 \times 0}}{{1 + 2}}} \right)$

$\Rightarrow \left( h,k \right)=\left( \frac{2x}{3},\frac{y}{3} \right)$

$\Rightarrow h=\frac{2x}{3}$ and $k = \frac{y}{3}$ 

$\Rightarrow x = \frac{{3h}}{2}$ and  y=3k

Therefore, the respective coordinates of A and B are $\left( \frac{3h}{2},0 \right)$ and $\left( {0,3k} \right)$ 

Now, the equation of the line AB passing through points $\left( {\frac{{3h}}{2},0} \right)$and$\left( {0,3k} \right)$is-

$\left( {y - 0} \right) = \frac{{3k - 0}}{{0 - \frac{{3h}}{2}}}\left( {x - \frac{{3h}}{2}} \right)$ 

$\Rightarrow y =  - \frac{{2k}}{h}\left( {x - \frac{{3h}}{2}} \right)$ 

$\Rightarrow hy=-2kx+3hk$ 

i.e. $2kx + hy = 3hk$ 

Thus, the required equation of a line is $2kx + hy = 3hk$.


20. By using the concept of the equation of a line, prove that the three points $\left( {3,0} \right),\left( { - 2, - 2} \right)$  and $\left( 8,2 \right)$ are collinear.

Ans:In order to show that the points $\left( {3,0} \right),\left( { - 2, - 2} \right)$ and$\left( {8,2} \right)$ are collinear, it sufficient to show that the line passing through points $\left( {3,0} \right)$  and $\left( -2,-2 \right)$ also passes through point$\left( {8,2} \right)$

Therefore the equation of the line passing through points$\left( {3,0} \right)$ and$\left( { - 2, - 2} \right)$ is-

$\left( {y - 0} \right) = \frac{{ - 2 - 0}}{{ - 2 - 3}}\left( {x - 3} \right)$

$\Rightarrow y=\frac{-2}{-5}\left( x-3 \right)$

$\Rightarrow 5y = 2x - 6$ 

i.e. $2x - 5y = 6$

It is observed that at $x=8$ and $y = 2$ 

L.H.S $= 2 \times 8 - 5 \times 2 = 16 - 10 = 6 =$R.H.S.

Therefore, the line passing through points $\left( 3,0 \right)$ and$\left( { - 2, - 2} \right)$ also passes through point$\left( {8,2} \right)$.

Hence, points $\left( {3,0} \right),\left( { - 2, - 2} \right)$ and$\left( {8,2} \right)$ are collinear.


NCERT Solutions for Class 11 Maths Chapters

 

NCERT Solution Class 11 Maths of Chapter 10 All Exercises

Chapter 10 - Straight Lines Exercises in PDF Format

Exercise 10.1

14 Questions & Solutions

Exercise 10.2

20 Questions & Solutions

Exercise 10.3

18 Questions & Solutions

Miscellaneous Exercise

24 Questions & Solutions


All Topics of Class 11 NCERT Maths for Exercise 10.2 

The topics and subtopics covered under exercise 10.2 Maths Class 11 NCERT are given below.

Section 10.3 deals with the various forms of the equation of a line. With the help of these equations, we can identify whether a point P(x0, y0) lies on the line or not. Equations of a line vary under different conditions.

Section 10.3.1 defines what are horizontal and vertical lines.

Section 10.3.2 is the point-slope form of the line. The line is basically non-vertical and non-horizontal. If we take two points on this line, say A (x0, y0) and B (x1, y1) then the equation of a line will be

y1 – y0 = m (x1 – x0)

Section 10.3.3 defines the equation of a line when three points on it are given. 

Section 10.3.4 defines the process of formation of the equation of the line of which slope and one intercept on the axis are given.

Section 10.3.5 shows the equation of the line in which two points are given. One point lies where the line cuts the X-axis and another is the point at Y-axis where the line intercepts.

If the line cut cuts X-axis at point a and Y-axis at point b, then the equation of the line are $ \frac{x}{a} + \frac{y}{b} = 1\,$

Section 10.3.6 defines the process of formation of line when two informations related to the line is given:

  1. Length of the perpendicular from the line to the origin.

  2. Angle between the X-axis and perpendicular from the origin to the line is given.


NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.2

Opting for the NCERT solutions for Ex 10.2 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.2 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.


Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 10 Exercise 10.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.


Besides these NCERT solutions for Class 11 Maths Chapter 10 Exercise 10.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

 

Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 10 Exercise 10.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well. 

Share this with your friends
SHARE
TWEET
SHARE
SUBSCRIBE