NCERT Solutions for Miscellaneous Exercise Chapter 9 Class 11 - Straight Lines

Question1: (a) Find the values of $k$ for which the line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is

Parallel to the x-axis.

Answer:

Given equation of line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
and equation of x-axis is $y=0$
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of $y=0$ is , $m^{\prime }=0$
and
Slope of line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is , $m=\frac{k-3}{4-k^2}$
Now,
$m=m^{\prime }$
$\frac{k-3}{4-k^2}=0$
$k-3=0$
$k=3$
Therefore, the value of k is 3

Question1: (b) Find the values of $k$ for which the line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is

Parallel to the y-axis.

Answer:

Given the equation of line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
and equation of y-axis is $x=0$
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of $y=0$ is , $m^{\prime }=\mathrm{\infty }=\frac{1}{0}$
and
Slope of line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is , $m=\frac{k-3}{4-k^2}$
Now,
$m=m^{\prime }$
$\frac{k-3}{4-k^2}=\frac{1}{0}$
$4-k^2=0$
$k=\pm 2$
Therefore, value of k is $\pm 2$

Question1: (c) Find the values of k for which the line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ is Passing through the origin.

Answer:

Given equation of line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
It is given that it passes through origin (0,0)
Therefore,
$(k-3).0-(4-k^2).0+k^2-7k+6=0$
$k^2-7k+6=0$
$k^2-6k-k+6=0$
$(k-6)(k-1)=0$
$k=6or1$
Therefore, value of k is $6or1$

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

Answer:

Let the intercepts on x and y-axis are a and b respectively
It is given that
$a+b=1anda.b=-6$
$a=1-b$
$\Rightarrow b.(1-b)=-6$
$\Rightarrow b-b^2=-6$
$\Rightarrow b^2-b-6=0$
$\Rightarrow b^2-3b+2b-6=0$
$\Rightarrow (b+2)(b-3)=0$
$\Rightarrow b=-2and3$
Now, when $b=-2\Rightarrow a=3$
and when $b=3\Rightarrow a=-2$
We know that the intercept form of the line is
$\frac{x}{a}+\frac{y}{b}=1$

Case (i) when a = 3 and b = -2
$\frac{x}{3}+\frac{y}{-2}=1$
$\Rightarrow 2x-3y=6$

Case (ii) when a = -2 and b = 3
$\frac{x}{-2}+\frac{y}{3}=1$
$\Rightarrow -3x+2y=6$
Therefore, equations of lines are $2x-3y=6and-3x+2y=6$

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

Answer:

Given the equation of the line is
$\frac{x}{3}+\frac{y}{4}=1$
we can rewrite it as
$4x+3y=12$
Let's take point on y-axis is $(0,y)$
It is given that the distance of the point $(0,y)$ from line $4x+3y=12$ is 4 units
Now,
$d=\left|\frac{A{x}_1+B{y}_1+C}{\sqrt{A^2+B^2}}\right|$
In this problem $A=4,B=3,C=-12,d=4and(x_1,y_1)=(0,y)$
$4=\left|\frac{4\times 0+3\times y-12}{\sqrt{4^2+3^2}}\right|=\left|\frac{3y-12}{\sqrt{16+9}}\right|=\left|\frac{3y-12}{5}\right|$

Case (i)

$4=\frac{3y-12}{5}$
$20=3y-12$
$y=\frac{32}{3}$
Therefore, the point is $(0,\frac{32}{3})$ -(i)

Case (ii)

$4=-\left(\frac{3y-12}{5}\right)$
$20=-3y+12$
$y=-\frac{8}{3}$
Therefore, the point is $(0,-\frac{8}{3})$ -(ii)
Therefore, points on the $y$-axis whose distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is $4$ units are $(0,\frac{32}{3})$ and $(0,-\frac{8}{3})$

Question 4: Find perpendicular distance from the origin to the line joining the points $(\cos \theta ,\sin \theta )$ and $(\cos \varphi ,\sin \varphi ).$.

Answer:

Equation of line passing through the points $(\cos \theta ,\sin \theta )$ and $(\cos \varphi ,\sin \varphi )$ is
$(y-\sin \theta )=\frac{\sin \varphi -\sin \theta }{\cos \varphi -\cos \theta }(x-\cos \theta )$
$\Rightarrow (\cos \varphi -\cos \theta )(y-\sin \theta )=(\sin \varphi -\sin \theta )(x-\cos \theta )$
$\Rightarrow y(\cos \varphi -\cos \theta )-\sin \theta (\cos \varphi -\cos \theta )=x(\sin \varphi -\sin \theta )-\cos \theta (\sin \varphi -\sin \theta )$
$\Rightarrow x(\sin \varphi -\sin \theta )-y(\cos \varphi -\cos \theta )=\cos \theta (\sin \varphi -\sin \theta )-\sin \theta (\cos \varphi -\cos \theta )$$\Rightarrow x(\sin \varphi -\sin \theta )-y(\cos \varphi -\cos \theta )=\sin (\theta -\varphi )$
$(\because \cos a\sin b-\sin a\cos b=\sin (a-b))$
Now, distance from origin(0,0) is
$d=\left|\frac{(\sin \varphi -\sin \theta ).0-(\cos \varphi -\cos \theta ).0-\sin (\theta -\varphi )}{\sqrt{(\sin \varphi -\sin \theta {)}^2+(\cos \varphi -\cos \theta {)}^2}}\right|$
$d=\left|\frac{-\sin (\theta -\varphi )}{\sqrt{({\sin }^2\varphi +{\cos }^2\varphi )+({\sin }^2\theta +{\cos }^2\theta )-2(\cos \theta \cos \varphi +\sin \theta \sin \varphi )}}\right|$
$d=\left|\frac{-\sin (\theta -\varphi )}{1+1-2\cos (\theta -\varphi )}\right|$ $(\because \cos a\cos b+\sin a\sin b=\cos (a-b)and{\sin }^{2}a+{\cos }^{2}a=1)$
$d=\left|\frac{-\sin (\theta -\varphi )}{2(1-\cos (\theta -\varphi ))}\right|$
$$\begin{gathered} d=\left|\frac{-2\sin \frac{\theta -\varphi }{2}\cos \frac{\theta -\varphi }{2}}{\sqrt{2.(2 \\ si{n}^2\frac{\theta -\varphi }{2})}}\right| \end{gathered}$$
$d=\left|\frac{-2\sin \frac{\theta -\varphi }{2}\cos \frac{\theta -\varphi }{2}}{2\sin \frac{\theta -\varphi }{2}}\right|$
$d=|\cos \frac{\theta -\varphi }{2}|$

Question 5: Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines $x-7y+5=0$ and $3x+y=0$.

Answer:

Point of intersection of the lines $x-7y+5=0$ and $3x+y=0$
$(-\frac{5}{22},\frac{15}{22})$
It is given that this line is parallel to y - axis i.e. $x=0$ which means their slopes are equal
Slope of $x=0$ is ,$m^{\prime }=\mathrm{\infty }=\frac{1}{0}$
Let the Slope of line passing through point $(-\frac{5}{22},\frac{15}{22})$ is m
Then,
$m=m^{\prime }=\frac{1}{0}$
Now, equation of line passing through point $(-\frac{5}{22},\frac{15}{22})$ and with slope $\frac{1}{0}$ is
$(y-\frac{15}{22})=\frac{1}{0}(x+\frac{5}{22})$
$x=-\frac{5}{22}$
Therefore, equation of line is $x=-\frac{5}{22}$

Question 6: Find the equation of a line drawn perpendicular to the line $\frac{x}{4}+\frac{y}{6}=1$ through the point, where it meets the $y$-axis.

Answer:

given equation of line is
$\frac{x}{4}+\frac{y}{6}=1$
we can rewrite it as
$3x+2y=12$
Slope of line $3x+2y=12$ , $m^{\prime }=-\frac{3}{2}$
Let the Slope of perpendicular line is m
$m=-\frac{1}{m^{\prime }}=\frac{2}{3}$
Now, the ponit of intersection of $3x+2y=12$ and $x=0$ is $(0,6)$
Equation of line passing through point $(0,6)$ and with slope $\frac{2}{3}$ is
$(y-6)=\frac{2}{3}(x-0)$
$3(y-6)=2x$
$2x-3y+18=0$
Therefore, equation of line is $2x-3y+18=0$

Question 7: Find the area of the triangle formed by the lines $y-x=0,x+y=0$ and $x-k=0$.

Answer:

Given equations of lines are
$y-x=0-(i)$
$x+y=0-(ii)$
$x-k=0-(iii)$
The point if intersection of (i) and (ii) is (0,0)
The point if intersection of (ii) and (iii) is (k,-k)
The point if intersection of (i) and (iii) is (k,k)
Therefore, the vertices of triangle formed by three lines are $(0,0),(k,-k)and(k,k)$
Now, we know that area of triangle whose vertices are $(x_1,y_1),(x_2,y_2)and(x_3,y_3)$ is
$A=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$
$A=\frac{1}{2}|0(-k-k)+k(k-0)+k(0+k)|$
$A=\frac{1}{2}|k^2+k^2|$
$A=\frac{1}{2}|2k^2|$
$A=k^2$
Therefore, area of triangle is $k^{2}squareunits$

Question 8: Find the value of $p$ so that the three lines $3x+y-2=0,px+2y-3=0$ and $2x-y-3=0$ may intersect at one point.

Answer:

Point of intersection of lines $3x+y-2=0$ and $2x-y-3=0$ is $(1,-1)$
Now, $(1,-1)$ must satisfy equation $px+2y-3=0$
Therefore,
$p(1)+2(-1)-3=0$
$p-2-3=0$
$p=5$
Therefore, the value of p is $5$

Question 9: If three lines whose equations are $y=m_{1}x+c_1,y=m_{2}x+c_2$ and $y=m_{3}x+c_3$ are concurrent, then show that $m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0$.

Answer:

Concurrent lines means they all intersect at the same point
Now, given equation of lines are
$y=m_{1}x+c_1-(i)$
$y=m_{2}x+c_2-(ii)$
$y=m_{3}x+c_3-(iii)$
Point of intersection of equation (i) and (ii) $(\frac{c_2-c_1}{m_1-m_2},\frac{m_1{c}_2-m_2{c}_1}{m_1-m_2})$

Now, lines are concurrent which means point $(\frac{c_2-c_1}{m_1-m_2},\frac{m_1{c}_2-m_2{c}_1}{m_1-m_2})$ also satisfy equation (iii)
Therefore,

$\frac{m_1{c}_2-m_2{c}_1}{m_1-m_2}=m_3.\left(\frac{c_2-c_1}{m_1-m_2}\right)+c_3$

$m_1{c}_2-m_2{c}_1=m_3(c_2-c_1)+c_3(m_1-m_2)$

$m_1(c_2-c_3)+m_2(c_3-c_1)+m_3(c_1-c_2)=0$

Hence proved

Question 10: Find the equation of the lines through the point $(3,2)$ which make an angle of ${45}^{\circ }$ with the line $x-2y=3$.

Answer:

Given the equation of the line is
$x-2y=3$
The slope of line $x-2y=3$ , $m_2=\frac{1}{2}$
Let the slope of the other line is, $m_1=m$
Now, it is given that both the lines make an angle ${45}^{\circ }$ with each other
Therefore,
$\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$
$\tan {45}^{\circ }=\left|\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\right|$
$1=\left|\frac{1-2m}{2+m}\right|$
Now,

Case (i)
$1=\frac{1-2m}{2+m}$
$2+m=1-2m$
$m=-\frac{1}{3}$
Equation of line passing through the point $(3,2)$ and with slope $-\frac{1}{3}$
$(y-2)=-\frac{1}{3}(x-3)$
$3(y-2)=-1(x-3)$
$x+3y=9-(i)$

Case (ii)
$1=-\left(\frac{1-2m}{2+m}\right)$
$2+m=-(1-2m)$
$m=3$
Equation of line passing through the point $(3,2)$ and with slope 3 is
$(y-2)=3(x-3)$
$3x-y=7-(ii)$

Therefore, equations of lines are $3x-y=7$ and $x+3y=9$

Question 11: Find the equation of the line passing through the point of intersection of the lines $4x+7y-3=0$ and $2x-3y+1=0$ that has equal intercepts on the axes.

Answer:

Point of intersection of the lines $4x+7y-3=0$ and $2x-3y+1=0$ is $(\frac{1}{13},\frac{5}{13})$
We know that the intercept form of the line is
$\frac{x}{a}+\frac{y}{b}=1$
It is given that line make equal intercepts on x and y axis
Therefore,
a = b
Now, the equation reduces to
$x+y=a$ -(i)
It passes through point $(\frac{1}{13},\frac{5}{13})$
Therefore,
$a=\frac{1}{13}+\frac{5}{13}=\frac{6}{13}$
Put the value of a in equation (i)
we will get
$13x+13y=6$
Therefore, equation of line is $13x+13y=6$

Question 12: Show that the equation of the line passing through the origin and making an angle $\theta$ with the line $y=mx+c$ is $\frac{y}{x}=\frac{m\pm \tan \theta }{1\mp m\tan \theta }$.

Answer:

Slope of line $y=mx+c$ is m
Let the slope of other line is m'
It is given that both the line makes an angle $\theta$ with each other
Therefore,
$\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$
$\tan \theta =\left|\frac{m-m^{\prime }}{1+m{m}^{\prime }}\right|$
$\mp (1+m{m}^{\prime })\tan \theta =(m-m^{\prime })$
$\mp \tan \theta +m^{\prime }(\mp m\tan \theta +1)=m$
$m^{\prime }=\frac{m\pm \tan \theta }{1\mp m\tan \theta }$
Now, equation of line passing through origin (0,0) and with slope $\frac{m\pm \tan \theta }{1\mp m\tan \theta }$ is
$(y-0)=\frac{m\pm \tan \theta }{1\mp m\tan \theta }(x-0)$
$\frac{y}{x}=\frac{m\pm \tan \theta }{1\mp m\tan \theta }$
Hence proved

Question 13: In what ratio, the line joining $(-1,1)$ and $(5,7)$ is divided by the line $x+y=4$ ?

Answer:

Equation of line joining $(-1,1)$ and $(5,7)$ is
$(y-1)=\frac{7-1}{5+1}(x+1)$
$\Rightarrow (y-1)=\frac{6}{6}(x+1)$
$\Rightarrow (y-1)=1(x+1)$
$\Rightarrow x-y+2=0$
Now, point of intersection of lines $x+y=4$ and $x-y+2=0$ is $(1,3)$
Now, let's suppose point $(1,3)$divides the line segment joining $(-1,1)$ and $(5,7)$ in $1:k$
Then,
$(1,3)=(\frac{k(-1)+1(5)}{k+1},\frac{k(1)+1(7)}{k+1})$
$1=\frac{-k+5}{k+1}and3=\frac{k+7}{k+1}$
$\Rightarrow k=2$
Therefore, the line joining $(-1,1)$ and $(5,7)$ is divided by the line $x+y=4$ in ratio $1:2$

Question 14: Find the distance of the line $4x+7y+5=0$ from the point $(1,2)$ along the line $2x-y=0$.

Answer:

point $(1,2)$ lies on line $2x-y=0$
Now, point of intersection of lines $2x-y=0$ and $4x+7y+5=0$ is $(-\frac{5}{18},-\frac{5}{9})$
Now, we know that the distance between two point is given by
$d=|\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}|$
$d=|\sqrt{(1+\frac{5}{18}{)}^2+(2+\frac{5}{9}{)}^2}|$
$d=|\sqrt{(\frac{23}{18}{)}^2+(\frac{23}{9}{)}^2}|$
$d=\left|\sqrt{\frac{529}{324}+\frac{529}{81}}\right|$
$d=\left|\sqrt{\frac{529+2116}{324}}\right|=\left|\sqrt{\frac{2645}{324}}\right|=\frac{23\sqrt{5}}{18}$
Therefore, the distance of the line $4x+7y+5=0$ from the point $(1,2)$ along the line $2x-y=0$ is $\frac{23\sqrt{5}}{18}units$

Question 15: Find the direction in which a straight line must be drawn through the point $(-1,2)$ so that its point of intersection with the line $x+y=4$ may be at a distance of $3$ units from this point.

Answer:

Let $(x_1,y_1)$ be the point of intersection
it lies on line $x+y=4$
Therefore,
$$\begin{gathered} x_1+y_1=4 \\ {x}_1=4-y_1-(i) \end{gathered}$$
Distance of point $(x_1,y_1)$ from $(-1,2)$ is 3
Therefore,
$3=|\sqrt{(x_1+1{)}^2+(y_1-2{)}^2}|$
Square both the sides and put value from equation (i)
$$\begin{gathered} 9=(5-y_1{)}^2+(y_1-2{)}^2 \\ 9=y_1^2+25-10y_1+y_1^2+4-4y_1 \\ 2y_1^2-14y_1+20=0 \\ {y}_1^2-7y_1+10=0 \\ {y}_1^2-5y_1-2y_1+10=0 \\ (y_1-2)(y_1-5)=0 \\ {y}_1=2or{y}_1=5 \end{gathered}$$
When $y_1=2\Rightarrow x_1=2$ point is $(2,2)$
and
When $y_1=5\Rightarrow x_1=-1$ point is $(-1,5)$
Now, slope of line joining point $(2,2)$ and $(-1,2)$ is
$m=\frac{2-2}{-1-2}=0$
Therefore, line is parallel to x-axis -(i)

or
slope of line joining point $(-1,5)$ and $(-1,2)$
$m=\frac{5-2}{-1+2}=\mathrm{\infty }$
Therefore, line is parallel to y-axis -(ii)

Therefore, line is parallel to x -axis or parallel to y-axis

Question 16: The hypotenuse of a right angled triangle has its ends at the points $(1,3)$ and $(-4,1)$ Find an equation of the legs (perpendicular sides) of the triangle.

Answer:

Slope of line OA and OB are negative times inverse of each other
Slope of line OA is , $m=\frac{3-y}{1-x}\Rightarrow (3-y)=m(1-x)$
Slope of line OB is , $-\frac{1}{m}=\frac{1-y}{-4-x}\Rightarrow (x+4)=m(1-y)$
Now,
for a given value of m we get these equations
If $m=\mathrm{\infty }$
$1-x=0and1-y=0$
$x=1andy=1$

Question 17: Find the image of the point $(3,8)$ with respect to the line $x+3y=7$ assuming the line to be a plane mirror.

Answer:

Let point $(a,b)$ is the image of point $(3,8)$ w.r.t. to line $x+3y=7$
line $x+3y=7$ is perpendicular bisector of line joining points $(3,8)$ and $(a,b)$
Slope of line $x+3y=7$ , $m^{\prime }=-\frac{1}{3}$
Slope of line joining points $(3,8)$ and $(a,b)$ is , $m=\frac{8-b}{3-a}$
Now,
$m=-\frac{1}{m^{\prime }}(\because linesareperpendicular)$
$\frac{8-b}{3-a}=3$
$8-b=9-3a$
$3a-b=1-(i)$
Point of intersection is the midpoint of line joining points $(3,8)$ and $(a,b)$
Therefore,
Point of intersection is $(\frac{3+a}{2},\frac{b+8}{2})$
Point $(\frac{3+a}{2},\frac{b+8}{2})$ also satisfy the line $x+3y=7$
Therefore,
$\frac{3+a}{2}+3.\frac{b+8}{2}=7$
$a+3b=-13-(ii)$
On solving equation (i) and (ii) we will get
$(a,b)=(-1,-4)$
Therefore, the image of the point $(3,8)$ with respect to the line $x+3y=7$ is $(-1,-4)$

Question 18: If the lines $y=3x+1$ and $2y=x+3$ are equally inclined to the line $y=mx+4$ , find the value of $m$.

Answer:

Given equation of lines are
$y=3x+1-(i)$
$2y=x+3-(ii)$
$y=mx+4-(iii)$
Now, it is given that line (i) and (ii) are equally inclined to the line (iii)
Slope of line $y=3x+1$ is , $m_1=3$
Slope of line $2y=x+3$ is , $m_2=\frac{1}{2}$
Slope of line $y=mx+4$ is , $m_3=m$
Now, we know that
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
Now,
$\tan {\theta }_1=\left|\frac{3-m}{1+3m}\right|$ and $\tan {\theta }_2=\left|\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\right|$

It is given that $\tan {\theta }_1=\tan {\theta }_2$
Therefore,
$\left|\frac{3-m}{1+3m}\right|=\left|\frac{1-2m}{2+m}\right|$
$\frac{3-m}{1+3m}=\pm \left(\frac{1-2m}{2+m}\right)$
Now, if $\frac{3-m}{1+3m}=\left(\frac{1-2m}{2+m}\right)$
Then,
$(2+m)(3-m)=(1-2m)(1+3m)$
$6+m-m^2=1+m-6m^2$
$5m^2=-5$
$m=\sqrt{-1}$
Which is not possible
Now, if $\frac{3-m}{1+3m}=-\left(\frac{1-2m}{2+m}\right)$
Then,
$(2+m)(3-m)=-(1-2m)(1+3m)$
$6+m-m^2=-1-m+6m^2$
$7m^2-2m-7=0$
$m=\frac{-(-2)\pm \sqrt{(-2{)}^2-4\times 7\times (-7)}}{2\times 7}=\frac{2\pm \sqrt{200}}{14}=\frac{1\pm 5\sqrt{2}}{7}$

Therefore, the value of m is $\frac{1\pm 5\sqrt{2}}{7}$

Question 19: If the sum of the perpendicular distances of a variable point $P(x,y)$ from the lines $x+y-5=0$ and $3x-2y+7=0$ is always $10$. Show that $P$ must move on a line.

Answer:

Given the equation of line are
$x+y-5=0-(i)$
$3x-2y+7=0-(ii)$
Now, perpendicular distances of a variable point $P(x,y)$ from the lines are

$d_1=\left|\frac{1.x+1.y-5}{\sqrt{1^2+1^2}}\right|$ $d_2=\left|\frac{3.x-2.y+7}{\sqrt{3^2+2^2}}\right|$
$d_1=\left|\frac{x+y-5}{\sqrt{2}}\right|$ $d_2=\left|\frac{3x-2y+7}{\sqrt{13}}\right|$
Now, it is given that
$d_1+d_2=10$
Therefore,
$\frac{x+y-5}{\sqrt{2}}+\frac{3x-2y+7}{\sqrt{13}}=10$
$(assumingx+y-5>0and3x-2y+7>0)$
$(x+y-5)\sqrt{13}+(3x-2y+7)\sqrt{2}=10\sqrt{26}$

$x(\sqrt{13}+3\sqrt{2})+y(\sqrt{13}-2\sqrt{2})=10\sqrt{26}+5\sqrt{13}-7\sqrt{2}$

Which is the equation of the line
Hence proved