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

Question:1(i) Reduce the following equations into slope - intercept form and find their slopes and the $y$ - intercepts.

$x+7y=0$

Answer:

Given equation is
$x+7y=0$
we can rewrite it as
$y= -\frac{1}{7}x$ -(i)
Now, we know that the Slope-intercept form of the line is
$y = mx+C$ -(ii)
Where m is the slope and C is some constant
On comparing equation (i) with equation (ii)
we will get
$m =- \frac{1}{7}$ and $C = 0$
Therefore, slope and y-intercept are $-\frac{1}{7} \ and \ 0$ respectively

Question:1(ii) Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.

$6x+3y-5=0$

Answer:

Given equation is
$6x+3y-5=0$
we can rewrite it as
$y= -\frac{6}{3}x+\frac{5}{3}\Rightarrow y = -2x+\frac{5}{3}$ -(i)
Now, we know that the Slope-intercept form of line is
$y = mx+C$ -(ii)
Where m is the slope and C is some constant
On comparing equation (i) with equation (ii)
we will get
$m =- 2$ and $C = \frac{5}{3}$
Therefore, slope and y-intercept are $-2 \ and \ \frac{5}{3}$ respectively

Question:1(iii) Reduce the following equations into slope - intercept form and find their slopes and the y - intercepts.

$y=0.$

Answer:

Given equation is
$y=0$ -(i)
Now, we know that the Slope-intercept form of the line is
$y = mx+C$ -(ii)
Where m is the slope and C is some constant
On comparing equation (i) with equation (ii)
we will get
$m =0$ and $C = 0$
Therefore, slope and y-intercept are $0 \ and \ 0$ respectively

Question:2(i) Reduce the following equations into intercept form and find their intercepts on the axes.

$3x+2y-12=0$

Answer:

Given equation is
$3x+2y-12=0$
we can rewrite it as
$\frac{3x}{12}+\frac{2y}{12} = 1$
$\frac{x}{4}+\frac{y}{6} = 1$ -(i)
Now, we know that the intercept form of line is
$\frac{x}{a}+\frac{y}{b} = 1$ -(ii)
Where a and b are intercepts on x and y axis respectively
On comparing equation (i) and (ii)
we will get
a = 4 and b = 6
Therefore, intercepts on x and y axis are 4 and 6 respectively

Question:2(ii)Reduce the following equations into intercept form and find their intercepts on the axes.

$4x-3y=6$

Answer:

Given equation is
$4x-3y=6$
we can rewrite it as
$\frac{4x}{6}-\frac{3y}{6} = 1$
$\frac{x}{\frac{3}{2}}-\frac{y}{2} = 1$ -(i)
Now, we know that the intercept form of line is
$\frac{x}{a}+\frac{y}{b} = 1$ -(ii)
Where a and b are intercepts on x and y axis respectively
On comparing equation (i) and (ii)
we will get
$a = \frac{3}{2}$ and $b = -2$
Therefore, intercepts on x and y axis are $\frac{3}{2}$ and -2 respectively

Question:2(iii) Reduce the following equations into intercept form and find their intercepts on the axes.

$3y+2=0$

Answer:

Given equation is
$3y+2=0$
we can rewrite it as
$y = \frac{-2}{3}$
Therefore, intercepts on y-axis are $\frac{-2}{3}$
and there is no intercept on x-axis

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

Answer:

Given the equation of the line is
$12(x+6)=5(y-2)$
we can rewrite it as
$12x+72=5y-10$
$12x-5y+82=0$
Now, we know that
$d= \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$ where A and B are the coefficients of x and y and C is some constant and $(x_1,y_1)$ is point from which we need to find the distance
In this problem A = 12 , B = -5 , c = 82 and $(x_1,y_1)$ = (-1 , 1)
Therefore,
$d = \frac{|12.(-1)+(-5).1+82|}{\sqrt{12^2+(-5)^2}} = \frac{|-12-5+82|}{\sqrt{144+25}}=\frac{|65|}{\sqrt{169}}=\frac{65}{13}= 5$
Therefore, the distance of the point $(-1,1)$ from the line $12(x+6)=5(y-2)$ is 5 units

Question:4 Find the points on the x-axis, whose distances from the line $\frac{x}{3}+\frac{y}{4}=1$ are $4$ units.

Answer:

Given equation of line is
$\frac{x}{3}+\frac{y}{4}=1$
we can rewrite it as
$4x+3y-12=0$
Now, we know that
$d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$
In this problem A = 4 , B = 3 C = -12 and d = 4
point is on x-axis therefore $(x_1,y_1)$ = (x ,0)
Now,
$4= \frac{|4.x+3.0-12|}{\sqrt{4^2+3^2}}= \frac{|4x-12|}{\sqrt{16+9}}= \frac{|4x-12|}{\sqrt{25}}= \frac{|4x-12|}{5}$
$20=|4x-12|\\ 4|x-3|=20\\ |x-3|=5$
Now if x > 3
Then,
$|x-3|=x-3\\ x-3=5\\ x = 8$
Therefore, point is (8,0)
and if x < 3
Then,
$|x-3|=-(x-3)\\ -x+3=5\\ x = -2$
Therefore, point is (-2,0)
Therefore, the points on the x-axis, whose distances from the line $\frac{x}{3}+\frac{y}{4}=1$ are $4$ units are (8 , 0) and (-2 , 0)

Question:5(i) Find the distance between parallel lines $15x+8y-34=0$ and $15x+8y+31=0$.

Answer:

Given equations of lines are
$15x+8y-34=0$ and $15x+8y+31=0$
it is given that these lines are parallel
Therefore,
$d = \frac{ |C_2-C_1|}{\sqrt{A^2+B^2}}$
$A = 15 , B = 8 , C_1= -34 \ and \ C_2 = 31$
Now,
$d = \frac{|31-(-34)|}{\sqrt{15^2+8^2}}= \frac{|31+34|}{\sqrt{225+64}}= \frac{|65|}{\sqrt{289}} = \frac{65}{17}$
Therefore, the distance between two lines is $\frac{65}{17} \ units$

Question:5(ii) Find the distance between parallel lines $l(x+y)+p=0$ and $l(x+y)-r = 0$

Answer:

Given equations of lines are
$l(x+y)+p=0$ and $l(x+y)-r = 0$
it is given that these lines are parallel
Therefore,
$d = \frac{ |C_2-C_1|}{\sqrt{A^2+B^2}}$
$A = l , B = l , C_1= -r \ and \ C_2 = p$
Now,
$d = \frac{|p-(-r)|}{\sqrt{l^2+l^2}}= \frac{|p+r|}{\sqrt{2l^2}}= \frac{|p+r|}{\sqrt{2}|l|}$
Therefore, the distance between two lines is $\frac{1}{\sqrt2}\left | \frac{p+r}{l} \right |$

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

Answer:

It is given that line is parallel to line $3x-4y+2=0$ which implies that the slopes of both the lines are equal
we can rewrite it as
$y = \frac{3x}{4}+\frac{1}{2}$
The slope of line $3x-4y+2=0$ = $\frac{3}{4}$
Now, the equation of the line passing through the point $(-2,3)$ and with slope $\frac{3}{4}$ is
$(y-3)=\frac{3}{4}(x-(-2))$
$4(y-3)=3(x+2)$
$4y-12=3x+6$
$3x-4y+18= 0$
Therefore, the equation of the line is $3x-4y+18= 0$

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

Answer:

It is given that line is perpendicular to the line $x-7y+5=0$
we can rewrite it as
$y = \frac{x}{7}+\frac{5}{7}$
Slope of line $x-7y+5=0$ ( m' ) = $\frac{1}{7}$
Now,
The slope of the line is $m = \frac{-1}{m'} = -7 \ \ \ \ \ \ \ \ \ \ \ (\because lines \ are \ perpendicular)$
Now, the equation of the line with $x$intercept $3$ i.e. (3, 0) and with slope -7 is
$(y-0)=-7(x-3)$
$y = -7x+21$
$7x+y-21=0$

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

Answer:

Given equation of lines are
$\sqrt{3}x+y=1$ and $x+\sqrt{3}y=1$

Slope of line $\sqrt{3}x+y=1$ is, $m_1 = -\sqrt3$

And
Slope of line $x+\sqrt{3}y=1$ is , $m_2 = -\frac{1}{\sqrt3}$

Now, if $\theta$ is the angle between the lines
Then,

$\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |$

$\tan \theta = \left | \frac{-\frac{1}{\sqrt3}-(-\sqrt3)}{1+(-\sqrt3).\left ( -\frac{1}{\sqrt3} \right )} \right | = \left | \frac{\frac{-1+3}{\sqrt3}}{1+1} \right |=| \frac{1}{\sqrt3}|$

$\tan \theta = \frac{1}{\sqrt3} \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \tan \theta = -\frac{1}{\sqrt3}$

$\theta = \frac{\pi}{6}=30^\circ \ \ \ \ \ \ \ or \ \ \ \ \ \ \theta =\frac{5\pi}{6}=150^\circ$

Therefore, the angle between the lines is $30^\circ \ and \ 150^\circ$

Question:9 The line through the points $(h,3)$ and $(4,1)$ intersects the line $7x-9y-19=0$ at right angle. Find the value of $h$.

Answer:

Line passing through points ( h ,3) and (4 ,1)

Therefore,Slope of the line is

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

$m =\frac{3-1}{h-4}$

This line intersects the line $7x-9y-19=0$ at right angle
Therefore, the Slope of both the lines are negative times inverse of each other
Slope of line $7x-9y-19=0$ , $m'=\frac{7}{9}$
Now,
$m=-\frac{1}{m'}$
$\frac{2}{h-4}= -\frac{9}{7}$
$14=-9(h-4)$
$14=-9h+36$
$-9h= -22$
$h=\frac{22}{9}$
Therefore, the value of h is $\frac{22}{9}$

Question:10 Prove that the line through the point $(x_1,y_1)$ and parallel to the line $Ax+By+C=0$ is $A(x-x_1)+B(y-y_1)=0.$

Answer:

It is given that line is parallel to the line $Ax+By+C=0$
Therefore, their slopes are equal
The slope of line $Ax+By+C=0$ , $m'= \frac{-A}{B}$
Let the slope of other line be m
Then,
$m =m'= \frac{-A}{B}$
Now, the equation of the line passing through the point $(x_1,y_1)$ and with slope $-\frac{A}{B}$ is
$(y-y_1)= -\frac{A}{B}(x-x_1)$
$B(y-y_1)= -A(x-x_1)$
$A(x-x_1)+B(y-y_1)= 0$
Hence proved

Question:11 Two lines passing through the point $(2,3)$ intersects each other at an angle of $60^{\circ}$. If slope of one line is $2$, find equation of the other line.

Answer:

Let the slope of two lines are $m_1 \ and \ m_2$ respectively
It is given the lines intersects each other at an angle of $60^{\circ}$ and slope of the line is 2
Now,
$m_1 = m\ and \ m_2= 2 \ and \ \theta = 60^\circ$
$\tan \theta = \left | \frac{m_2-m_1}{1+m_1m_2} \right |$
$\tan 60^\circ = \left | \frac{2-m}{1+2m} \right |$
$\sqrt3 = \left | \frac{2-m}{1+2m} \right |$
$\sqrt3 = \frac{2-m}{1+2m} \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \sqrt 3 = -\left ( \frac{2-m}{1+2m} \right )$
$m = \frac{2-\sqrt3}{2\sqrt3+1} \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ m = \frac{-(2+\sqrt3)}{2\sqrt3-1}$
Now, the equation of line passing through point (2 ,3) and with slope $\frac{2-\sqrt3}{2\sqrt3+1}$ is
$(y-3)= \frac{2-\sqrt3}{2\sqrt3+1}(x-2)$
$(2\sqrt3+1)(y-3)=(2-\sqrt3)(x-2)$
$x(\sqrt3-2)+y(2\sqrt3+1)=-1+8\sqrt3$ -(i)

Similarly,
Now , equation of line passing through point (2 ,3) and with slope $\frac{-(2+\sqrt3)}{2\sqrt3-1}$ is
$(y-3)=\frac{-(2+\sqrt3)}{2\sqrt3-1}(x-2)$
$(2\sqrt3-1)(y-3)= -(2+\sqrt3)(x-2)$
$x(2+\sqrt3)+y(2\sqrt3-1)=1+8\sqrt3$ -(ii)

Therefore, equation of line is $x(\sqrt3-2)+y(2\sqrt3+1)=-1+8\sqrt3$ or $x(2+\sqrt3)+y(2\sqrt3-1)=1+8\sqrt3$

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

Answer:

Right bisector means perpendicular line which divides the line segment into two equal parts
Now, lines are perpendicular which means their slopes are negative times inverse of each other
Slope of line passing through points $(3,4)$ and $(-1,2)$ is
$m'= \frac{4-2}{3+1}= \frac{2}{4}=\frac{1}{2}$
Therefore, Slope of bisector line is
$m = - \frac{1}{m'}= -2$
Now, let (h , k) be the point of intersection of two lines
It is given that point (h,k) divides the line segment joining point $(3,4)$ and $(-1,2)$ into two equal part which means it is the mid point
Therefore,
$h = \frac{3-1}{2} = 1\ \ \ and \ \ \ k = \frac{4+2}{2} = 3$
$(h,k) = (1,3)$
Now, equation of line passing through point (1,3) and with slope -2 is
$(y-3)=-2(x-1)\\ y-3=-2x+2\\ 2x+y=5$
Therefore, equation of line is $2x+y=5$

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

Answer:

Let suppose the foot of perpendicular is $(x_1,y_1)$
We can say that line passing through the point $(x_1,y_1) \ and \ (-1,3)$ is perpendicular to the line $3x-4y-16=0$
Now,
The slope of the line $3x-4y-16=0$ is , $m' = \frac{3}{4}$
And
The slope of the line passing through the point $(x_1,y_1) \ and \ (-1,3)$is, $m = \frac{y-3}{x+1}$
lines are perpendicular
Therefore,
$m = -\frac{1}{m'}\\ \frac{y_1-3}{_1+1} = -\frac{4}{3}\\ 3(y_1-3)=-4(x_1+1)\\ 4x_1+3y_1$

$=5 -(i)$
Now, the point $(x_1,y_1)$ also lies on the line $3x-4y-16=0$
Therefore,
$3x_1-4y_1=16 \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii)
we will get
$x_1 = \frac{68}{25} \ and \ y_1 =-\frac{49}{25}$
Therefore, $(x_1,y_1) = \left ( \frac{68}{25},-\frac{49}{25} \right )$

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

Answer:

We can say that line passing through point $(0,0) \ and \ (-1,2)$ is perpendicular to line $y=mx+c$
Now,
The slope of the line passing through the point $(0,0) \ and \ (-1,2)$ is , $m = \frac{2-0}{-1-0}= -2$
lines are perpendicular
Therefore,
$m = -\frac{1}{m'} = \frac{1}{2}$ - (i)
Now, the point $(-1,2)$ also lies on the line $y=mx+c$
Therefore,
$2=\frac{1}{2}.(-1)+C\\ C = \frac{5}{2} \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Therefore, the value of m and C is $\frac{1}{2} \ and \ \frac{5}{2}$ respectively

Question:15 If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x\cos \theta -y\sin \theta =k\cos 2\theta$ and $x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k$ , respectively, prove that $p^2+4q^2=k^2$.

Answer:

Given equations of lines are $x\cos \theta -y\sin \theta =k\cos 2\theta$ and $x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k$

We can rewrite the equation $x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k$ as

$x\sin \theta +y\cos \theta = k\sin\theta\cos\theta$
Now, we know that

$d = \left | \frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}} \right |$

In equation $x\cos \theta -y\sin \theta =k\cos 2\theta$

$A= \cos \theta , B = -\sin \theta , C = - k\cos2\theta \ and \ (x_1,y_1)$

$= (0,0)$

$p= \left | \frac{\cos\theta .0-\sin\theta.0-k\cos2\theta }{\sqrt{\cos^2\theta+(-\sin\theta)^2}} \right | = |-k\cos2\theta|$
Similarly,
in the equation $x\sin \theta +y\cos \theta = k\sin\theta\cos\theta$

$A= \sin \theta , B = \cos \theta , C = -k\sin\theta\cos\theta \ and \ (x_1,y_1)$

$= (0,0)$

$q= \left | \frac{\sin\theta .0+\cos\theta.0-k\sin\theta\cos\theta }{\sqrt{\sin^2\theta+\cos^2\theta}} \right | = |-k\sin\theta\cos\theta|= \left | -\frac{k\sin2\theta}{2} \right |$
Now,

$p^2+4q^2=(|-k\cos2\theta|)^2+4.(|-\frac{k\sin2\theta}{2})^2$

$= k^2\cos^22\theta+4.\frac{k^2\sin^22\theta}{4}$
$=k^2(\cos^22\theta+\sin^22\theta)$
$=k^2$
Hence proved

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

Answer:

Let suppose foot of perpendicular is $(x_1,y_1)$
We can say that line passing through point $(x_1,y_1) \ and \ A(2,3)$ is perpendicular to line passing through point $B(4,-1) \ and \ C(1,2)$
Now,
Slope of line passing through point $B(4,-1) \ and \ C(1,2)$ is , $m' = \frac{2+1}{1-4}= \frac{3}{-3}=-1$
And
Slope of line passing through point $(x_1,y_1) \ and \ (2,3)$ is , $m$
lines are perpendicular
Therefore,
$m = -\frac{1}{m'}= 1$
Now, equation of line passing through point (2 ,3) and slope with 1
$(y-3)=1(x-2)$
$x-y+1=0$ -(i)
Now, equation line passing through point $B(4,-1) \ and \ C(1,2)$ is
$(y-2)=-1(x-1)$
$x+y-3=0$
Now, perpendicular distance of (2,3) from the $x+y-3=0$ is
$d= \left | \frac{1\times2+1\times3-3}{\sqrt{1^2+1^2}} \right |= \left | \frac{2+3-3}{\sqrt{1+1}} \right |= \frac{2}{\sqrt{2}}=\sqrt2$ -(ii)

Therefore, equation and length of the line is $x-y+1=0$ and $\sqrt2$ respectively

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

Answer:

we know that intercept form of line is
$\frac{x}{a}+\frac{y}{b} = 1$
we know that
$d = \left | \frac{Ax_1+bx_2+C}{\sqrt{A^2+B^2}} \right |$
In this problem
$A = \frac{1}{a},B = \frac{1}{b}, C =-1 \ and \ (x_1,y_1)= (0,0)$
$p= \left | \frac{\frac{1}{a}\times 0+\frac{1}{b}\times 0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \right | = \left | \frac{-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \right |$
On squaring both the sides
we will get
$\frac{1}{p^2}= \frac{1}{a^2}+\frac{1}{b^2}$
Hence proved