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# NCERT Solutions for Class 11 Maths Chapter 9 - Straight Lines Exercise 9.3

Last updated date: 19th Jul 2024
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## NCERT Solutions for Excercise 9.3 Class 11 Maths Chapter 9 Straight Lines - FREE PDF Download

In Class 11 Maths NCERT Solutions for Exercise 9.3 of Chapter 9 on Straight Lines, we explore various forms of the equations of lines and their applications. Class 11 Ex 9.3 focuses on important concepts such as the distance of a point from a line, the distance between two parallel lines, and the angle between two intersecting lines.

Table of Content
1. NCERT Solutions for Excercise 9.3 Class 11 Maths Chapter 9 Straight Lines - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 9 Exercise 9.3 Class 11 | Vedantu
3. Formulas Used in Class 11 Ex 9.3
4. Access NCERT Solutions for Maths Class 11 Chapter 9 - Straight Lines
5. Class 11 Maths Chapter 9: Exercises Breakdown
6. CBSE Class 11 Maths Chapter 9 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

Students will learn how to apply the distance formula to find how far a point is from a given line and use the properties of parallel lines to calculate the distance between them. Additionally, understanding how to find the angle between two lines will enhance your ability to analyze the geometric relationships between lines.  Access the latest CBSE Class 11 Maths Syllabus here.

## Glance on NCERT Solutions Maths Chapter 9 Exercise 9.3 Class 11 | Vedantu

• The slope-intercept form explains the converting equations of lines into the form y = mx + b (where m is the slope and b is the y-intercept) and finding the slope and y-intercept from the equation.

• Parallel and perpendicular lines are used for Identifying parallel and perpendicular lines based on their slopes.

• The distance formula is used to Calculate the distance between a point and a line.

• The angle between two lines determines the angle of intersection between two lines using their slopes.

• Learn to rewrite equations of lines in slope-intercept form.

• Determine if two lines are parallel, perpendicular, or neither based on their slopes.

• Ex 9.3 Class 11 contains 17 Questions and Solutions.

## Formulas Used in Class 11 Ex 9.3

• Slope-intercept form: y = mx + b

• Distance formula: Distance between a point (x₁, y₁) and a line Ax + By + C = 0 is: |Ax₁ + By₁ + C| / √(A² + B²) (assuming A and B are not both zero)

Competitive Exams after 12th Science

## 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 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, 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 vertexA=y-x=1. Length of AD= Length of the perpendicular from A (2,3) toBC. 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

## 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.

## FAQs on NCERT Solutions for Class 11 Maths Chapter 9 - Straight Lines Exercise 9.3

1. What are the topics and subtopics of Class 11 Maths Chapter 9- Straight Lines?

The topics and subtopics of Class 11 Maths Chapter 9 titled Straight Lines are given below. Take a look.

• 9 - Straight Lines

• 9.1 - Introduction

• 9.2 - Slope of Line

• 9.3 - Various Forms of the Equation of Line

• 9.4 - General Equation of Line

• 9.5 - Distance of a Point From a Line

2. What does Exercise 9.3 of NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines deal with?

NCERT Solutions are provided to help the students understand the steps to solve mathematical problems that are provided in the textbook. The subject matter experts at Vedantu stick to the syllabus while preparing the solutions. The problem-solving method provided in the examples is followed while preparing the NCERT Solutions for class 11 as well.

The Exercise 9.3 of NCERT Solutions for Class 11 Maths Chapter 9- Straight Lines is based on the following topics:

1. General Equation of a Line

1. Different forms of Ax + By + C = 0

2. Distance of a Point From a Line

1. Distance between two parallel lines

3. How many questions are there in the Exercise 9.3 of NCERT Solutions for Class 11 Maths Chapter 9 - Straight Lines?

17 questions are there in total in the Exercise 9.3 of NCERT Solutions for Class 11 Maths Chapter 9 - Straight Lines.

4. Can I get access to the PDF format of the Exercise 9.3 of NCERT Solutions for Class 11 Maths Chapter 9 - Straight Lines?

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.3 (Ex 9.3) is available on Vedantu site along with all chapter exercises at one place. These solutions are prepared by the expert teachers as per CBSE curriculum and guidelines. Class 11 Maths Chapter 9 Straight Lines Exercise 9.3 questions with solutions to help you to revise the entire syllabus and score the best possible marks.

5. What all things are included in Class 11 Maths Exercise 9.3?

Chapter 9 Exercise 9.3 NCERT Solutions for Class 11 Maths Straight Lines explains how algebraic equations can be used to learn about a straight line's trajectory. There are numerous possibilities for a straight line, such as being parallel to the X or Y axis or going through the origin. When we look at how the equation of such a line is simplified when some of the terms in the equation become zero, we can identify some of these circumstances that make the straight line a little more distinct.

6. How does Algebra provide insight into the concepts of a straight line?

Simply looking at the simplified equation y= MX or y=0 or x=0 gives us an instant knowledge of the line's specialization. The questions in this NCERT Solutions Class 11 Mathematics Chapter 9 exercise 9.3 show how Algebra may help us understand the principles of a straight line by using its various forms. Taking notes from Vedantu will help students in solving these questions easily. Then some questions use line equations to assist us to discover the relationship between a group of lines.

7. How can you find perpendicular and parallel lines?

We can determine if they are parallel or perpendicular and if they intersect, where they intersect. Some questions incorporate trigonometry, and we can see how by combining different concepts, we may create an outstanding tool for analysis. In Class 11 Maths NCERT Solutions In Chapter 9, Exercise 9.3, there are 17 questions, seven of which are brief and eleven of which are extended answer-type questions. Vedantu provides every concept in the form of a PDF too, which you can download free of cost.

8. How can students solve the 9.3 Exercise?

Class 11 Maths NCERT Solutions Chapter 9 Exercise 9.3 Straight Lines focuses on utilizing various slope shapes and identifying key elements associated with them. Thus, memorizing the formula for the general form of slopes and understanding how to apply it to estimate various characteristics of a straight line is the key to successfully answering these issues. The Vedantu website have all the questions on the website in which you will find all the solutions of the exercise present in the NCERT book.

9. How can imagining different values help in solving this exercise easily?

The inclination should be to read the problem and visualize how it will be plotted on the axis when doing the NCERT solutions Class 11 Maths Chapter 9 Exercise 9.3 Straight Lines. Visualizing how it will behave for various values of the variables while focusing on the unique conditions might also aid in the learning process. On Vedantu, you can find the basic concepts of the chapter which will help you in understanding the topic in depth. Solving numerical will be easier if you know the concept behind it.

10. What is the significance of the distance formula in coordinate geometry?

As we studied in Class 11 Ex 9.3 NCERT Solutions, The distance formula is used to calculate the shortest distance from a point to a line, which is essential in many geometric applications, such as finding the nearest point on a line or optimizing distances in various contexts.

11. Why is it important to understand the distance between parallel lines which we discussed in Ex 9.3 Class 11 Maths NCERT Solutions?

Understanding the distance between parallel lines helps in solving problems related to parallelism and ensuring that structures and shapes maintain consistent spacing, which is crucial in fields like engineering and architecture.