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Get stepwise NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 – Straight Lines. Download FREE PDF for accurate answers and quick exam revision, based on the latest CBSE 2025-26 syllabus.

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CBSE Class 11 Mathematics Chapter 10 Straight Lines – NCERT Solutions 2025-26

Are you aiming to master NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 before your board exams? This part of coordinate geometry deepens your understanding of straight lines through slope, angle, intercept and general forms, all mapped to the latest syllabus. Each exercise builds the logic and technique needed for exam-perfect answers, making these solutions an essential step in your revision journey.

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Exercise 10.3 focuses on transforming equations, solving for concurrency, and applying the distance formula, which often appear in the “class 11 maths straight lines exercise 10.3” section of CBSE papers. With chapter weightage reaching up to 12 marks in final exams, solving these questions connects textbook theory to real exam patterns seamlessly.


You’ll find step-by-step guidance and all board-approved formats curated by Vedantu for clarity. Practicing with complete solutions and validating your approach boosts confidence and ensures your answers match official keys. For ongoing syllabus updates, always refer to the Class 11 Maths syllabus as your reference.

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Access NCERT Solutions for Maths Class 11 Chapter 9 - Straight Lines

Exercise 9.3

1. Reduce the following equation into slope-intercept form and find their slopes and the $y$ intercepts.

i. $x  +  7y  =  0$ 

Ans: The equation is $x+7y=0$.

We can write it as $y  =  -  \dfrac{1}{7}  x  +  0$$\ldots (1)$

This equation is of the form $y=mx+c$, where $m=-\dfrac{1}{7}$ and $c=0$ 

Therefore, the equation $\left( 1 \right)$ is the slope-intercept form, where the slope and the y-intercept are $-\dfrac{1}{7}$ and $0$ respectively.

ii. $6x+3y-5=0$ 

Ans: The given equation is $6x+3y-5=0$. 

We can write it as $y=\dfrac{1}{3}(-6x+5)$

$\Rightarrow y=-2x+\dfrac{5}{3}\quad \ldots (1)$

This equation is of the form$\text{y}=\text{mx}+\text{c}$, where $\text{m}=-2$ and$\text{c}=\dfrac{5}{3}$. Therefore, equation $\left( 1 \right)$  in the slope-intercept form, where the slope and the $y$-intercept are $-2$ and $\dfrac{5}{2}$ respectively.

iii. $y=0$

Ans: The given equation is $y=0$. 

We can write it as $y=0.x+0$$.....\left( 1 \right)$

This equation is of the form$y=mx+c$, where $m=0$ and $c=0$. Therefore, equation $\left( 1 \right)$ is in the slope-intercept form, where the slope and the $y$-intercept are \[0\] and \[0\] respectively.


2. Reduce the following equations into intercept form and find their intercepts on the axes.

i. $3x+2y-12=0$

Ans: The given equation is $3x+2y-12=0$.

We can write it as,

$3x+2y=12$

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

i.e. $\dfrac{x}{4}+\dfrac{y}{6}-1\quad \ldots (1)$

This equation is of the form$\dfrac{x}{a}+\dfrac{y}{b}=1$, where $a=4$ and$b=6$. 

Therefore, equation (1) is in the intercept form, where the intercepts on the $\text{x}$ and $\text{y}$ axes are $4$ and $6$ respectively.

ii. $4x-3y=6$

Ans:  The given equation is $4x-3y=6$. 

We can write it as,

$\dfrac{4x}{6}-\dfrac{3y}{6}=1$

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

i.e. $\dfrac{x}{\left( \dfrac{3}{2} \right)}+\dfrac{y}{(-2)}=1$$\ldots (2)$

Therefore, equation $(2)$ is in the intercept form, where the intercepts on $\text{x}$ and $\text{y}$ axes are $\dfrac{3}{2}$ and $-2$ respectively.

iii. $3y+2=0$.

Ans: The given equation is $3\text{y}+2=0$. 

We can write it as $3y=-2$

i.e. $\dfrac{\text{y}}{\left( -\dfrac{2}{3} \right)}=1$$\ldots (3)$

Therefore, equation is in the$\dfrac{x}{a}+\dfrac{y}{b}=1$, where $a=0$ and$b=-\dfrac{2}{3}$. 

Therefore, equation (3) is in the intercept form, where the intercept on the $y$-axis is $-\dfrac{2}{3}$ .

 It has no intercept on the $x$-axis.


3. Find the distance of the points $(-1,1)$ from the line $12(x+6)=5(y-2)$.

Ans: The equation of the line is$12(x+6)=5(y-2)$. 

$\Rightarrow 12x+72=5y-10$

$\Rightarrow 12x-5y+82=0$

Compare the equation $\left( 1 \right)$ with  the general equation of the line $Ax+By+C=0$, we obtain $A=12$, $\text{B}=-5$, and $C=82$.

The perpendicular distance (d) of a line $Ax+By+C=0$ from a point $\left( {{x}_{1}},{{y}_{1}} \right)$ is  

\[\text{d}=\dfrac{\left| \text{A}{{\text{x}}_{1}}+B{{y}_{1}}+\text{C} \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\]

The given point is$\left( {{x}_{1}},{{y}_{1}} \right)=(-1,1)$. 

So, the distance of the point $(-1,1)$ from the given line

$=\frac{|12(-1)+(-5)(1)+82|}{\sqrt{(12)^{2}+(-5)^{2}}}$ units $=\frac{|-12-5+82|}{\sqrt{169}}$ units $=\frac{|65|}{13}$ units $=5$ units


4. Find the points on the x-axis at what distance from the line $\dfrac{x}{3}+\dfrac{y}{4}=1$ is $4$ units.

Ans: The given equation of the line is $\dfrac{x}{3}+\dfrac{y}{4}=1$

Or $4x+3y-12=0$

Compare the equation (1) with the general equation of the line$Ax+By+C=0$.

We get,

$A=4$,$B=3$, and $C=-12$

Let $(a,0)$ be the point on the x-axis whose distance from the given line is $4$units.

The perpendicular distance $(d)$ of a line $Ax+By+C=0$ from a point $\left( {{x}_{1}},{{y}_{1}} \right)$ is $d=\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+c \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$

Then $4 = \frac{{\left| {4a + 3 \times 0 - 12} \right|}}{{\sqrt {{4^2} + {3^2}} }}$

$ \Rightarrow {\text{\; }}4 =\frac{{\left| {4a - 12} \right|}}{5}$

$ \Rightarrow {\text{\; }}|4a - 12|= 20$

$ \Rightarrow {\text{\; }} \pm (4a -12) = 20$

$\Rightarrow (4a - 12) = 20$ or $ - (4a - 12) = 20$

$ \Rightarrow 4a = 20 + 12$

or $ - 4a + 12 = 20$

$ \Rightarrow 4a = 32\,or\,\,4a = 12 - 20$

$ \Rightarrow a = \frac{{32}}{4}\,or\,\,a = \frac{{ - 8}}{4}$

$ \Rightarrow a = 8$ or $ - 2$

Thus, the required points on $x$the -axis are $(-2,0)$ and $(8,0)$.


5. Find the distance between parallel lines

i. $15x+8y-34=0$ and $15x+8y+31=0$

Ans: The distance $\left( d \right)$between parallel lines $Ax  +  By  +{{C}_{1}}  =  0$ and $\text{Ax}+\text{By}+{{\text{C}}_{2}}=0$ is$d=\dfrac{\left| {{C}_{1}}-{{C}_{2}} \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$.

The parallel lines are $15x+8y-34=0$ and $15x+8y+31=0$ .

Here, 

$A=15,~\text{B}=8,{{C}_{1}}=-34$, and${{C}_{2}}=31$.

Therefore, the distance between the parallel lines is

\[d  =  \dfrac{\left| {{C}_{1}}  -  {{C}_{2}} \right|}{\sqrt{{{A}^{2}}  +  {{B}^{2}}}}  =  \dfrac{\left| -34  -31 \right|}{\sqrt{{{15}^{2}}  +  {{8}^{2}}}}  \]

\[=  \dfrac{\left| -65 \right|}{\sqrt{289}}  \]Units.

\[=  \dfrac{65}{17}  \]Units.

ii. $l(x+y)+p=0$ and $l(x+y)-r-0$

Ans: The distance $\left( d \right)$between parallel lines $Ax  +  By  +{{C}_{1}}  =  0$ and $\text{Ax}+\text{By}+{{\text{C}}_{2}}=0$ is given by,

$d=\dfrac{\left| {{C}_{1}}-{{C}_{2}} \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$.

The parallel lines are $l(\text{x}+y)+\text{p}=0$ and $l(\text{x}+\text{y})-\text{r}=0$

i.e $lx+ly+p=0$ and $lx+ly-r=0$

Here,

$\text{A}=l ,\text{B}=l,{{C}_{1}}=p$, and ${{C}_{2}}=-{r}$

Therefore, the distance between the parallel lines is

\[d  =  \dfrac{\left| {{C}_{1}}  -  {{C}_{2}} \right|}{\sqrt{{{A}^{2}}  +  {{B}^{2}}}}  =  \dfrac{\left| p  +  r \right|}{\sqrt{{{l}^{2}}  +  {{l}^{2}}}}  \]\[=  \dfrac{\left| p  +  r \right|}{\sqrt{2 {{l}^{2}}}}  \]Units.

\[=  \dfrac{\left| p  +  r \right|}{l\sqrt{ 2}}  \]Units.

\[\Rightarrow d  =  \dfrac{1}{\sqrt{2}}  \dfrac{\left| p  +  r \right|}{l}  \]Units.


6. Find the equation of the line parallel to the line $3x-4y+z=0$ and passing through the point$\left( -  2,3 \right)$.

Ans: The equation of the line is given as,

$3x-4y+2=0$

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

or $y=\dfrac{3}{4}x+\dfrac{1}{2}$ 

Which is of the form $y=mx+c$

$\therefore $ The slope of the given line $=\dfrac{3}{4}$

 It is known that parallel lines have the same slope.

$\therefore $ The slope of the other line $=m=\dfrac{3}{4}$

Equation of line having slope m and passing through $\left( {{x_1},{y_1}} \right)$is given by

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

Now, the equation of the line that has a slope of $\dfrac{3}{4}$ and passes through the points $(-2,3)$ is

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

$\Rightarrow 4y-12=3x+6$

i.e ,$3x-4y+18=0$


7. Find the equation of the line perpendicular to the line $x-7y+5=0$ and having $x$ intercept $3$.

Ans: The equation of the line is $x-7y+5=0$.

Or $y=\dfrac{1}{7}x+\dfrac{5}{7}$,

 which is of the form $y=mx+c$

$\therefore $ The slope of the given line $=\dfrac{1}{7}$ .

The slope of the line perpendicular to the line having a slope of $\dfrac{1}{7}$ is $m=-\dfrac{1}{\left( \dfrac{1}{7} \right)}=-7$.

 The equation of the line with slope $-7$ and $x$-intercept 3 is given by $y=m(x-d)$

$\Rightarrow y=-7(x-3)$

$\Rightarrow y=-7x+21$

$\Rightarrow 7x+y=21$


8. Find angles between the lines $\sqrt{3}x+y=1$ and $x+\sqrt{3}y=1$.

Ans: The given lines are $\sqrt{3}x+y=1$ and $x+\sqrt{3}y=1$

$y=-\sqrt{3x}+1 \quad -(1)$  and $y=-\dfrac{1}{\sqrt{3}}x+\dfrac{1}{\sqrt{3}}\quad -(2)$

The slope of the line (1) is ${{m}_{1}}=-\sqrt{3}$, while the slope of the line (2) is ${{m}_{2}}=-\dfrac{1}{\sqrt{3}}$.

The acute angle  between the two lines is given by

$\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+m,{{m}_{2}}} \right|$

$\tan \theta =\left| \dfrac{-\sqrt{3}+\dfrac{1}{\sqrt{3}}}{1+(-\sqrt{3})\left( -\dfrac{1}{\sqrt{3}} \right)} \right|$

$\tan \theta =\left| \dfrac{\dfrac{-3+1}{\sqrt{3}}}{1+1} \right|=\left| \dfrac{-2}{2\times \sqrt{3}\mid } \right|$

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

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

Thus, the angle between the given lines is either ${{30}^{{}^\circ }}$ or ${{180}^{*}}-{{30}^{{}^\circ }}={{150}^{{}^\circ }}$.


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

Ans: The slope of the line passing through points $(\text{h},3)$ and $(4,1)$ is

${m_1} = \frac{{\left( {{y_2} - {y_1}} \right)}}{{\left( {{x_2} - {x_1}} \right)}}\, = \,\frac{{1 - 3}}{{4 - h}} = \frac{{ - 2}}{{4 - h}}$

The slope of the line $7\text{x}-9y-19=0$ or $y=\dfrac{7}{9}x-\dfrac{19}{9}$ is ${{\text{m}}_{2}}=\dfrac{7}{9}$.

It is given that the two lines are perpendicular. 

$\therefore \Rightarrow \text{  }{{m}_{1}}\times {{m}_{2}}=-1$

$\Rightarrow \left( {\frac{{ - 2}}{{4 - h}}} \right) \times \frac{7}{9}$

$\Rightarrow \dfrac{-14}{36-9h}=-1$

$\Rightarrow \text{  }14=36-9h$

$\Rightarrow 9h=36-14$

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

Thus, the value  $h$ is $\dfrac{22}{9}$


10. Prove that the line through the point $\left( {{x}_{1}},{{y}_{1}} \right)$ and parallel to the line $Ax+By+C=0$ is. $A\left( x-{{x}_{1}} \right)+B\left( y-{{y}_{1}} \right)=0$

Ans: The slope of the line $Ax+By+C=0$ or $y=\left( \dfrac{-A}{B} \right)x+\left( \dfrac{-C}{B} \right)$ is $m=-\dfrac{A}{B}$ .

The parallel lines have the same slope.

$\therefore $ The slope of the other line $m=-\dfrac{A}{B}$

The equation of the line passing through a point $\left( {{x}_{1}},{{y}_{1}} \right)$ and having a slope $m=-\dfrac{A}{B}$ is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-{{y}_{1}}=-\dfrac{A}{B}\left( x-{{x}_{1}} \right)$

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

$A\left( x-{{x}_{1}} \right)+B\left( y-{{y}_{1}} \right)=0$

Hence, the line through point $\left( {{x}_{1}}-{{y}_{1}} \right)$ and parallel to line $\text{Ax}+\text{By}+C=0$ is $A \left( x-{{x}_{1}} \right)+B\left( y-{{y}_{1}} \right)=0$


 11. Two lines passing through the points $(2,3)$ intersect each other at an angle

 of $60{}^\circ $. If the slope of the one line is $2$, find the equation of the other line.

Ans: It is given that the slope of the first line, ${{m}_{1}}=2$. 

Let the slope of the other line be ${{m}_{2}}$ .

The angle between the two lines is ${{60}^{{}^\circ }}$.

$\therefore \tan 60{}^\circ =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$

$\Rightarrow \sqrt{3}=\left| \dfrac{2-{{m}_{2}}}{1+2{{m}_{2}}} \right|$

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

$\Rightarrow \sqrt{3}=\dfrac{2-{{m}_{2}}}{1+2{{m}_{2}}}  \text{or}  \sqrt{3}=-\left( \dfrac{2-{{m}_{3}}}{1+2{{m}_{2}}} \right)$

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

\[\Rightarrow \sqrt{3}+2\sqrt{3}{{m}_{2}}+{{\text{m}}_{2}}=2\] or \[\sqrt{3}+2\sqrt{3}{{m}_{2}}-{{m}_{2}}  =  -2\]

$\Rightarrow \sqrt{3}+(2\sqrt{3}+1){{\text{m}}_{2}}=2$or $\sqrt{3}+(2\sqrt{3}-1){{\text{m}}_{3}}=-2$

$\Rightarrow {{m}_{2}}=\dfrac{2-\sqrt{3}}{(2\sqrt{3}+1)}  \text{or}  {{m}_{2}}=\dfrac{-(2+\sqrt{3})}{(2\sqrt{3}-1)}$

Case 1: $\quad {{\text{m}}_{2}}=\left( \dfrac{2-\sqrt{3}}{(2\sqrt{3}+1)} \right)$

The equation of the line passing through the point $(2,3)$ and having a slope of $\dfrac{2-\sqrt{3}}{(2\sqrt{3}+1)}$is 

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

$(2\sqrt{3}+1)y-3(2\sqrt{3}+1)=(2-\sqrt{3})x-(2-\sqrt{3})2$

$(\sqrt{3}-2)x+(2\sqrt{3}+1)y=-4+2\sqrt{3}+6\sqrt{3}+3$

$(\sqrt{3}-2)x+(2\sqrt{3}+1)y=-1+8\sqrt{3}$

In this case, the equation of the other line is $(\sqrt{3}-2)x+(2\sqrt{3}+1)y=-1+8\sqrt{3}$.

Case 2: $\quad {{\text{m}}_{2}}=\left( \dfrac{-\left( 2+\sqrt{3} \right)}{(2\sqrt{3}-1)} \right)$

The equation of the line passing through the point $(2,3)$ and having a slope of $\left( \dfrac{-\left( 2+\sqrt{3} \right)}{(2\sqrt{3}-1)} \right)$ is 

$(y - 3) = \frac{{ - (2 + \sqrt 3 )}}{{(2\sqrt 3  - 1)}}(x - 2)$

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

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

$\left( {2+ \sqrt 3 } \right)x + \left( {2\sqrt 3 - 1} \right)y = 4 + 2\sqrt 3  +6\sqrt 3  - 3$

$\left( {2 + \sqrt 3 } \right)x + \left( {2\sqrt 3  - 1} \right)y = 1 + 8\sqrt 3 $

In this case, the equation of the other line is$(2+\sqrt{3})x+(2\sqrt{3}-1)y=1+8\sqrt{3}$

The required equation of the line is  $(\sqrt{3}-2)x+(2\sqrt{3}+1)y=-1+8\sqrt{3}$ or $(2+\sqrt{3})x+(2\sqrt{3}-1)y=1+8\sqrt{3}$


12. Find the equation of the right bisector of the line segment joining the points $(3,4)$ and $(-1,2)$.

Ans: The right bisector of a line segment bisects the line segment at $90{}^\circ $.

 The end-point $A(3,4)$ and $B(-1,2)$of the line segment .

Accordingly, the mid-point of $AB=\left( \dfrac{3-1}{2},\dfrac{4+2}{0} \right)-(1,3)$ 

The Slope of $\text{AB}=-\dfrac{2-4}{-1-3}=\dfrac{-2}{-4}=\dfrac{1}{2}$

$\therefore $ The slope of the line perpendicular to $AB=-\dfrac{1}{\left( \dfrac{1}{2} \right)}=-2$ 

The equation of the line passing through $(1,3)$ and having a slope of $-2$ is $(y-3)=-2(x-1)$

$\Rightarrow y-3=-2x+2$

$\Rightarrow 2x+y=5$

Thus, the required equation of the line is $2x+y=5$.


13. Find the coordinates of the foot of the perpendicular from the points $(-1,3)$ to the line $3x-4y-16=0$

Ans: Let $(a,b)$ be the coordinates of the foot of the perpendicular from the points $(-1,3)$ to the line $3x-4y-16=0$.


the coordinates of the foot of the perpendicular


The slope of the line joining $(-1,3)$ and $(a,b)$, 

$\Rightarrow \text{  }$${{m}_{1}}=\dfrac{b-3}{a+1}$ 

Slope of the line $3x-4y-16=0$ or $y=\dfrac{3}{4}x-4,{{m}_{2}}=\dfrac{3}{4}$ ,

 The above two lines are perpendicular, at, ${{\text{m}}_{1}}{{\text{m}}_{2}}=-1$ 

$\therefore \Rightarrow \text{  }\left( \dfrac{b-3}{a+1} \right)\times \left( \dfrac{3}{4} \right)=-1$

$\Rightarrow \dfrac{3b-9}{4a+4}=-1$

$\Rightarrow 3b-9=-4a-4$

$\Rightarrow 4a+3b=5\quad .....(1)$

The point $(a$, b) lies on the line $3x-4y=16$.

$\therefore \Rightarrow \text{  }3a-4b=16.....(2)$

Solve the equations $(1)$ and (2), we get,

 $a=\dfrac{68}{25}$ and $b=-\dfrac{49}{25}$

Thus, the required coordinates of the foot of the perpendicular are $\left( \dfrac{68}{25},-\dfrac{49}{25} \right)$


14. The perpendicular from the origin to the fine $y=mx+c$ meets it at the point $(-1,2)$. Find the values of $m$ and $c$.

Ans: The  equation of the line is $y=mx+c$. 

The perpendicular from the origin meets the given line at $(-1,2)$. 

So, the line joining the points $(0,0)$ and $(-1,2)$ is perpendicular to the given line. 

$\therefore $ The slope of the line joining $(0,0)$ and $(-1,2)=\dfrac{2}{-1}=-2$ The slope of the given line is $\text{m}$.

 $\therefore \text{m}\times -2=-1\quad $     The two  lines are perpendicular

$\Rightarrow m=\dfrac{1}{2}$.

The  points $(-1,2)$ lie on the given line, it satisfies the equation $y=mx+c$

$\therefore 2=\text{m}(-1)+c$

$\Rightarrow 2=2+\dfrac{1}{2}(-1)+c$

 $\Rightarrow c=2+\dfrac{1}{2}=\dfrac{5}{2}$

The respective values of $\text{m}$ and $\text{c}$ are $\dfrac{1}{2}$ and $\dfrac{5}{2}$.


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

Ans: The equation of lines are $x\cos \theta -y\sin \theta =k\cos 2\theta \quad \ldots (1) $

$x\sec \theta +y$ cases $\theta =k\quad \ldots (2)$

The perpendicular distance (d) of a line $Ax+By+C=0$ from a point $\left( {{x}_{1}},{{x}_{2}} \right)$ is given by $d=\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+c \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$

Compare the  equation $\left( 1 \right)$  the general equation of line 

ie., $Ax+By+C=0$, we obtain $A=\cos \theta $ ,$B=\sin \theta $, and $C=k\cos 2\theta $.

$\text{p}$ is the length of the perpendicular from $(0,0)$ to line $\left( 1 \right)$. \[\therefore p=\dfrac{|A(0)+B(0)+C|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}=\dfrac{|C|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\]

$\therefore p=\dfrac{|-k\cos 2\theta |}{\sqrt{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }}=|-k\cos 2\theta |\quad \ldots (3)$

 Compare the  equation (2) to the general equation of line ie., $Ax+By+C=0$, we obtain $A=\sec \theta $ , $B=\operatorname {cosec} \theta $, and $C=-k$

It is given that $q$ is the length of the perpendicular from $(0,0)$ to line (2). 

$\therefore \mathrm{q}=\frac{|\mathrm{A}(\mathrm{O})+\mathrm{B}(\mathrm{O})+\mathrm{C}|}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}}=\frac{|\mathrm{C}|}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}}=\frac{|-\mathrm{k}|}{\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}}$

From (3) and (4), we have

$p^{2}+4 q^{2}-(|-k \cos 2 \theta|)^{2}+4\left(\frac{|-k|}{\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}}\right)^{2}$

$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\sec ^{2} \theta+\operatorname{cosec}^{2} \theta\right)}$

$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}\right)}$

$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta}\right)}$

$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\frac{1}{\sin ^{2} \theta \cos ^{2} \theta}\right)}$

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

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

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

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

$=\mathbf{k}^{2}$

Hence, vre proved that $\mathrm{p}^{2}+4 \mathrm{q}^{2}=\mathrm{k}^{2}$


16. In the triangle $\text{ABC}$ with vertices $\text{A}(2,3),\text{B}(4,-1)$ and $\text{C}(1,2)$, find the equation and length of altitude from the vertex $A$.

Ans: Let $\text{AD}$ be the altitude of the triangle $\text{ABC}$ from the vertex $\text{A}$. 

 $\Rightarrow \text{  }AD\bot BC$,


equation and length of altitude from the vertex A


The slope of the line BC $\dfrac[21][1-4]$=$-1$

Therefore slope of line AD  $= \dfrac{-1}{-1}$=$1$

The equation of the line AD passing through the point $A (2,3)$ and having a slope $1$ is $(y-3)=1(x-2)$

$\Rightarrow x-y+1=0$

$\Rightarrow y-x=1$

Therefore, equation of the altitude from a vertex$A=y-x=1$.

Length of $AD=$ Length of the perpendicular from A $(2,3)$ to$BC$.

The equation of $\text{BC}$ is 

$(y+1)=\dfrac{2+1}{1-4}(x-4)$

$\Rightarrow \text{  }(y+1)=-1(x-4)$

$\Rightarrow y+1=-x+4$

$\Rightarrow x+y-3=0$$\ldots (1)$

The perpendicular distance $(d)$ of a line $Ax+By+C=0$ from a point $\left( {{x}_{1}},{{y}_{1}} \right)$ is $d=\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+c \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$

 Compare the equation $(1)$ to the general equation of the line$Ax+By+C=0$.

We get,

 $A=1$,$\text{B}=1$ and $C=-3$

Length of AD $=  \dfrac{\left| 1\times 2+1\times 3-3 \right|}{\sqrt{{{1}^{2}}+{{1}^{2}}}}$ units 

$=\dfrac{\left| 2 \right|}{\sqrt{2}}$units $=\sqrt{2 }$units.

Length of AD $=\sqrt{2 }$units.

Thus, the equation and length of the altitude from vertex $A$ are $y-x=1$ and $\sqrt{2}$ wits respectively.


17. If $\text{p}$ is the length of the perpendicular from the origin to the line whose intercepts on the axes are $a$, and $\text{b}$, then show that $\dfrac{1}{{{p}^{2}}}=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}$.

Ans: The equation of a line whose intercepts on the axes are $a$ and $\text{b}$ is $\dfrac{x}{a}+\dfrac{y}{b}=1$

$\text{bx}+ay=\text{ab}$

Or $bx+ay-ab=0$

The perpendicular distance $(d)$ of a line $Ax+By+C=0$ from a point $\left( {{x}_{1}},{{y}_{1}} \right)$ is $d=\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+c \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$

Compare the equation $(1)$ with the general equation of the line$Ax+By+C=0$, 

We obtain$A=b$,$\text{B}=a$ and $C=-ab$

Therefore, if $p$ is the length of the perpendicular from a point $\left( {{x}_{1}},{{y}_{1}} \right)=(0,0)$ to lime$\left( 1 \right)$. We obtain $p  =  \dfrac{\left| A\left( 0 \right)  +  B\left( 0 \right)  -  ab \right|}{\sqrt{{{b}^{2}}+{{a}^{2}}}}$

$\Rightarrow p=\dfrac{|-ab|}{\sqrt{{{b}^{2}}+{{a}^{2}}}}$.

Square both sides, 

We get

$\Rightarrow \text{  }{{p}^{2}}  =  \dfrac{{{\left( -  ab \right)}^{2}}}{{{a}^{2}}  +  {{b}^{2}}}$

$\Rightarrow \text{  }{{p}^{2}}\left( {{a}^{2}}  +  {{b}^{2}} \right)  =  {{a}^{2}}{{b}^{2}}$

$\Rightarrow \text{  }\dfrac{{{a}^{2}}  +  {{b}^{2}}}{{{a}^{2}}{{b}^{2}}}  =  \dfrac{1}{{{p}^{2}}}$

$\Rightarrow \text{  }\dfrac{1}{{{p}^{2}}}  =  \dfrac{1}{{{a}^{2}}}  +  \dfrac{1}{{{b}^{2}}}$

Hence proved that  $\dfrac{1}{{{p}^{2}}}=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}$


Conclusion

Ex 9.3 Class 11 Maths NCERT Solutions has provided an in-depth understanding of several advanced concepts related to straight lines. You have learned how to calculate the distance of a point from a line, the distance between two parallel lines, and the angle between two intersecting lines. By understanding these topics covered in class 11 ex 9.3, you have enhanced your ability to solve complex geometric problems and gained a deeper appreciation for the properties of lines in coordinate geometry.


Class 11 Maths Chapter 9: Exercises Breakdown

Exercise

Number of Questions

Exercise 9.1

11 Questions & Solutions

Exercise 9.2

19 Questions & Solutions

Miscellaneous Exercise

23 Questions & Solutions


CBSE Class 11 Maths Chapter 9 Other Study Materials


Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


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FAQs on Get stepwise NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 – Straight Lines. Download FREE PDF for accurate answers and quick exam revision, based on the latest CBSE 2025-26 syllabus.

1. How do I download the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 PDF from Vedantu?

You can easily download the NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 PDF from Vedantu by following these steps:

  • Visit the official Vedantu solution page for Class 11 Maths Chapter 10 Exercise 10.3.
  • Look for a clearly marked download button or PDF link near the top of the page.
  • Click the link to get instant access to the complete, syllabus-mapped PDF for Straight Lines Exercise 10.3.
All solutions are prepared as per the latest CBSE guidelines and are perfect for exam preparation.

2. Are the solutions for Exercise 10.3 strictly as per the latest CBSE syllabus and exam pattern?

Yes, the solutions for Exercise 10.3 strictly follow the latest CBSE Class 11 Maths syllabus and exam pattern:

  • All steps are provided as per the most recent NCERT and CBSE marking scheme (2025 syllabus update).
  • The problem-solving approach and formats are designed for board exams, JEE, and NEET relevance.
  • Detailed explanations and correct formats ensure you won't lose marks for missing steps.

3. What is the easiest way to understand slope and intercept forms of a straight line?

The best way to understand slope and intercept forms is by learning their formulas, usage, and examples:

  • Slope-intercept form: y = mx + c, where m is slope and c is y-intercept. Use when slope and intercept are known.
  • Intercept form: x/a + y/b = 1, where a and b are x- and y-intercepts. Apply when both intercepts are given.
  • Practice board-style solved questions and visualise each form graphically for clarity.
These forms are foundational for Class 11 Coordinate Geometry.

4. Can I access solutions for other exercises and miscellaneous of Chapter 10 (Straight Lines)?

Yes, you can access NCERT Solutions for all exercises and miscellaneous questions of Class 11 Maths Chapter 10 – Straight Lines on Vedantu:

  • Complete, stepwise solutions for all sub-exercises including 10.1, 10.2, 10.3 and the Miscellaneous Exercise.
  • Quick links to each exercise and downloadable PDFs are provided on the main chapter page.
  • Additional resources like important questions and revision notes are also available for thorough exam preparation.

5. Does Vedantu provide tips/hacks for solving straight lines questions faster?

Absolutely! Vedantu includes exam-oriented tips and solving hacks for Straight Lines Class 11:

  • Quick formula boxes for all line forms (slope, intercept, normal, general).
  • Shortcuts and stepwise tricks for finding slopes, intercepts, and distances in MCQs.
  • Orange-highlighted revision points and error alerts to avoid common mistakes.
  • Last-minute revision guides and one-liner summaries for all key concepts.
These strategies boost both speed and accuracy for CBSE, JEE, and NEET exams.

6. Why should I use Vedantu’s NCERT solutions before my board or JEE/NEET exams?

Vedantu’s NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 give you an exam-ready edge because:

  • All answers are stepwise, board-format, and 100% syllabus-mapped.
  • Solutions are peer-reviewed by expert educators and regularly updated to match the latest CBSE/JEE patterns.
  • Each answer is detailed yet concise for maximum marks and quick revision.
  • Downloadable PDFs and revision notes make last-minute study easier.
This ensures clarity, accuracy and confidence ahead of your board and entrance exams.

7. What are the main forms of straight line equations covered in Exercise 10.3?

Exercise 10.3 covers all major forms of straight line equations:

  • Slope-Intercept Form: y = mx + c
  • Intercept Form: x/a + y/b = 1
  • Normal Form: x cos α + y sin α = p
  • General Form: Ax + By + C = 0
Understanding these is essential for solving most CBSE board and JEE questions on straight lines.

8. How can I quickly revise all important formulas for straight lines before exams?

Use a revision strategy focused on the most important straight lines formulas:

  • Prepare a formula sheet listing:
    • Slope-intercept form
    • Intercept form
    • Normal form
    • Slope between two points: m = (y₂–y₁)/(x₂–x₁)
    • Angle between two lines
    • Distance of a point from a line
  • Practice one-word and MCQ questions based on these formulas.
  • Refer to Vedantu's PDF formula sheet and summary tables for fast revision.

9. How do I solve problems on concurrent lines and collinearity in Exercise 10.3?

To solve concurrent lines and collinearity problems in Exercise 10.3:

  • For concurrency, set the equations so all three lines meet at a single point and solve the system.
  • To test collinearity, use the area method or see if the same equation/model is satisfied by all points.
  • Apply stepwise NCERT solutions for logical justification, as required by the CBSE.
  • Use provided solved examples to see official reasoning and calculation techniques.

10. Are these Class 11 Maths solutions available in Hindi medium as well?

Yes, NCERT Solutions for Class 11 Maths Chapter 10 (Straight Lines) in Hindi medium are available on Vedantu:

  • All solutions, including Exercise 10.3, are fully translated and syllabus-aligned.
  • PDF downloads for Hindi medium can be found beside the English medium downloads.
  • This ensures accessibility for both English and Hindi medium CBSE students.

11. Does Exercise 10.3 help in JEE and NEET preparation?

Yes, mastering Exercise 10.3 of Class 11 Maths Chapter 10 (Straight Lines) builds foundational skills required for JEE and NEET:

  • Covers all key line forms, transformations, and problem types commonly asked in competitive exams.
  • Enhances algebraic manipulation and logical reasoning skills.
  • High overlap with previous year JEE and NEET questions on straight lines and coordinate geometry.

12. What types of questions are most frequently asked from Class 11 Maths Chapter 10 Straight Lines in CBSE exams?

In CBSE exams, the most asked questions from Class 11 Maths Straight Lines are:

  • Finding the equation of a line in different forms
  • Calculating the slope, distance between points, and angle between lines
  • Proof-based or transformation questions (collinearity, concurrency)
  • Application of various equations to real-life or graphical problems
Practicing complete NCERT solutions for Exercise 10.3 prepares you thoroughly for all these question types.