NCERT Solutions for Class 11 Maths Chapter 9 Exercise 9.2 - Straight Lines

Question 2: Find the equation of the line which satisfy the given conditions:

Passing through the point $(-4,3)$ with slope $\frac{1}{2}$.

Answer:

We know that , equation of line passing through point $(x_1,y_1)$ and with slope m is given by
$(y-y_1)=m(x-x_1)$
Now, equation of line passing through point (-4,3) and with slope $\frac{1}{2}$ is
$(y-3)=\frac{1}{2}(x-(-4))\\ 2y-6=x+4\\ x-2y+10 = 0$
Therefore, equation of the line is $x-2y+10 = 0$

Question 3: Find the equation of the line which satisfy the given conditions:

Passing through $(0,0)$ with slope $m$.

Answer:

We know that the equation of the line passing through the point $(x_1,y_1)$ and with slope m is given by
$(y-y_1)=m(x-x_1)$
Now, the equation of the line passing through the point (0,0) and with slope m is
$(y-0)=m(x-0)\\ y = mx$
Therefore, the equation of the line is $y = mx$

Question 4: Find the equation of the line which satisfy the given conditions:

Passing through $(2,2\sqrt{3})$ and inclined with the x-axis at an angle of $75^{\circ}$.

Answer:

We know that the equation of the line passing through the point $(x_1,y_1)$ and with slope m is given by
$(y-y_1)=m(x-x_1)$
we know that
$m = \tan \theta$
where $\theta$ is angle made by line with positive x-axis measure in the anti-clockwise direction
$m = \tan75^\circ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \theta=75^\circ \ given)$
$m = \frac{\sqrt3+1}{\sqrt3-1}$

Now, the equation of the line passing through the point $(2,2 \sqrt{ } 3)$ and with slope $m=\frac{\sqrt{3+1}}{\sqrt{3}-1}$ is

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

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

$=4(\sqrt{3}-1)
$
Therefore, the equation of the line is $(\sqrt{3}+1) x-(\sqrt{3}-1) y=4(\sqrt{3}-1)$

Question 5: Find the equation of the line which satisfy the given conditions:

Intersecting the $x$-axis at a distance of $3$ units to the left of origin with slope $-2$.

Answer:

We know that the equation of the line passing through the point $(x_1,y_1)$ and with slope m is given by
$(y-y_1)=m(x-x_1)$
Line Intersecting the $x$-axis at a distance of $3$ units to the left of origin which means the point is (-3,0)
Now, the equation of the line passing through the point (-3,0) and with slope -2 is
$(y-0)= -2(x-(-3))\\ y = -2x-6\\ 2x+y+6=0$
Therefore, the equation of the line is $2x+y+6=0$

Question 6: Find the equation of the line which satisfy the given conditions:

Intersecting the $y$-axis at a distance of $2$ units above the origin and making an angle of $30^{\circ}$

with positive direction of the x-axis.

Answer:

We know that , equation of line passing through point $(x_1,y_1)$ and with slope m is given by
$(y-y_1)=m(x-x_1)$
Line Intersecting the y-axis at a distance of 2 units above the origin which means point is (0,2)
we know that
$m = \tan \theta\\ m = \tan 30^\circ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \theta = 30 ^\circ \ given)\\ m = \frac{1}{\sqrt3}$
Now, the equation of the line passing through the point (0,2) and with slope $\frac{1}{\sqrt3}$ is
$(y-2)= \frac{1}{\sqrt3}(x-0)\\ \sqrt3(y-2)= x\\ x-\sqrt3y+2\sqrt3=0$
Therefore, the equation of the line is $x-\sqrt3y+2\sqrt3=0$

Question 7: Find the equation of the line which satisfy the given conditions:

Passing through the points $(-1,1)$ and $(2,-4)$.

Answer:

We know that , equation of line passing through point $(x_1,y_1)$ and with slope m is given by
$(y-y_1)=m(x-x_1)$
Now, it is given that line passes throught point (-1 ,1) and (2 , -4)
$m = \frac{y_2-y_1}{x_2-x_1}\\ \\ m = \frac{-4-1}{2+1}= \frac{-5}{3}$
Now, equation of line passing through point (-1,1) and with slope $\frac{-5}{3}$ is
$(y-1)= \frac{-5}{3}(x-(-1))\\ \\3(y-1)=-5(x+1)\\ 3y-3$

$=-5x-5\\ 5x+3y+2=0$

Question 8: The vertices of $\Delta \hspace{1mm}PQR$ are $P(2,1),Q(-2,3)$ and $R(4,5)$. Find equation of the median through the vertex $R$.

Answer:

The vertices of $\Delta \hspace{1mm}PQR$ are $P(2,1),Q(-2,3)$ and $R(4,5)$
Let m be RM b the median through vertex R
Coordinates of M (x, y ) = $\left ( \frac{2-2}{2},\frac{1+3}{2} \right )= (0,2)$
Now, slope of line RM
$m = \frac{y_2-y_1}{x_2-x_1} = \frac{5-2}{4-0}= \frac{3}{4}$
Now, equation of line passing through point $(x_1,y_1)$ and with slope m is
$(y-y_1)= m(x-x_1)$
equation of line passing through point (0 , 2) and with slope $\frac{3}{4}$ is
$(y-2)= \frac{3}{4}(x-0)\\ \\ 4(y-2)=3x\\ 4y-8$

$=3x\\ 3x-4y+8=0$
Therefore, equation of median is $3x-4y+8=0$

Question 9: Find the equation of the line passing through $(-3,5)$ and perpendicular to the line through the points $(2,5)$ and $(-3,6)$.

Answer:

It is given that the line passing through $(-3,5)$ and perpendicular to the line through the points $(2,5)$ and $(-3,6)$
Let the slope of the line passing through the point (-3,5) is m and
Slope of line passing through points (2,5) and (-3,6)
$m' = \frac{6-5}{-3-2}= \frac{1}{-5}$
Now this line is perpendicular to line passing through point (-3,5)
Therefore,
$m= -\frac{1}{m'} = -\frac{1}{\frac{1}{-5}}= 5$

Now, equation of line passing through point $(x_1,y_1)$ and with slope m is
$(y-y_1)= m(x-x_1)$
equation of line passing through point (-3 , 5) and with slope 5 is
$(y-5)= 5(x-(-3))\\ \\ (y-5)=5(x+3)\\ y-5$

$=5x+15\\ 5x-y+20=0$
Therefore, equation of line is $5x-y+20=0$

Question 10: A line perpendicular to the line segment joining the points $(1,0)$ and $(2,3)$ divides it in the ratio $1:n$. Find the equation of the line.

Answer:

Co-ordinates of point which divide line segment joining the points $(1,0)$ and $(2,3)$ in the ratio $1:n$ is
$\left ( \frac{n(1)+1(2)}{1+n},\frac{n.(0)+1.(3)}{1+n} \right )= \left ( \frac{n+2}{1+n},\frac{3}{1+n} \right )$
Let the slope of the perpendicular line is m
And Slope of line segment joining the points $(1,0)$ and $(2,3)$ is
$m'= \frac{3-0}{2-1}= 3$
Now, slope of perpendicular line is
$m = -\frac{1}{m'}= -\frac{1}{3}$
Now, equation of line passing through point $(x_1,y_1)$ and with slope m is
$(y-y_1)= m(x-x_1)$
equation of line passing through point $\left ( \frac{n+2}{1+n},\frac{3}{1+n} \right )$ and with slope $-\frac{1}{3}$ is
$(y- \frac{3}{1+n})= -\frac{1}{3}(x- (\frac{n+2}{1+n}))\\ 3y(1+n)-9$

$=-x(1+n)+n+2\\ x(1+n)+3y(1+n)=n+11$
Therefore, equation of line is $x(1+n)+3y(1+n)=n+11$

Question 11: Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point $(2,3)$.

Answer:

Let (a, b) are the intercept on x and y-axis respectively
Then, the equation of the line is given by
$\frac{x}{a}+\frac{y}{b}= 1$
Intercepts are equal which means a = b
$\frac{x}{a}+\frac{y}{a}= 1\\ \\ x+y = a$
Now, it is given that line passes through the point (2,3)
Therefore,
$a = 2+ 3 = 5$
therefore, equation of the line is $x+ y = 5$

Question 12: Find equation of the line passing through the point $(2,2)$ and cutting off intercepts on the axes whose sum is $9$.

Answer:

Let (a, b) are the intercept on x and y axis respectively
Then, the equation of line is given by
$\frac{x}{a}+\frac{y}{b}= 1$
It is given that
a + b = 9
b = 9 - a
Now,
$\frac{x}{a}+\frac{y}{9-a } = 1\\ \\ x(9-a)+ay= a(9-a)\\ 9x-ax+ay$

$=9a-a^2$
It is given that line passes through point (2 ,2)
So,
$9(2)-2a+2a=9a-a^2\\ a^2-9a+18$

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

$=0\\ (a-6)(a-3)= 0\\ a=6 \ \ or \ \ a = 3$

case (i) a = 6 b = 3
$\frac{x}{6}+\frac{y}{3}= 1\\ \\ x+2y = 6$

case (ii) a = 3 , b = 6
$\frac{x}{3}+\frac{y}{6}= 1\\ \\ 2x+y = 6$
Therefore, equation of line is 2x + y = 6 , x + 2y = 6

Question 13: Find equation of the line through the point $(0,2)$ making an angle $\frac{2\pi }{3}$ with the positive $x$-axis.

Also, find the equation of line parallel to it and crossing the $y$-axis at a distance of $2$ units below the origin.

Answer:

We know that
$m = \tan \theta \\ m = \tan \frac{2\pi}{3} = -\sqrt3$
Now, equation of line passing through point (0 , 2) and with slope $-\sqrt3$ is
$(y-2)= -\sqrt3(x-0)\\ \sqrt3x+y-2=0$
Therefore, equation of line is $\sqrt3x+y-2=0$ -(i)

Now, It is given that line crossing the $y$-axis at a distance of $2$ units below the origin which means coordinates are (0 ,-2)
This line is parallel to above line which means slope of both the lines are equal
Now, equation of line passing through point (0 , -2) and with slope $-\sqrt3$ is
$(y-(-2))= -\sqrt3(x-0)\\ \sqrt3x+y+2=0$
Therefore, equation of line is $\sqrt3x+y+2=0$

Question 14: The perpendicular from the origin to a line meets it at the point $(-2,9)$ , find the equation of the line.

Answer:

Let the slope of the line is m
and slope of a perpendicular line is which passes through the origin (0, 0) and (-2, 9) is
$m' = \frac{9-0}{-2-0}= \frac{9}{-2}$
Now, the slope of the line is
$m = -\frac{1}{m'}= \frac{2}{9}$
Now, the equation of line passes through the point (-2, 9) and with slope $\frac{2}{9}$ is
$(y-9)=\frac{2}{9}(x-(-2))\\ \\ 9(y-9)$

$=2(x+2)\\ 2x-9y+85 = 0$
Therefore, the equation of the line is $2x-9y+85 = 0$

Question 15: The length $L$(in centimetre) 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 $C$

Answer:

It is given that
If $C=20$ then $L=124.942$
and If $C=110$ then $L=125.134$
Now, if we assume C along the x-axis and L along y-axis
Then, we will get coordinates of two points (20 , 124.942) and (110 , 125.134)
Now, the relation between C and L is given by equation
$(L-124.942)= \frac{125.134-124.942}{110-20}(C-20)$
$(L-124.942)= \frac{0.192}{90}(C-20)$
$L= \frac{0.192}{90}(C-20)+124.942$
Which is the required relation

Question 16: The owner of a milk store finds that, he can sell 980 litres of milk each week at $Rs\hspace{1mm}14/litre$ and $1220$ litres of milk each week at $Rs\hspace{1mm}16/litre$ . Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at $Rs\hspace{1mm}17/litre$

Answer:

It is given that the owner of a milk store sell
980 litres milk each week at $Rs\hspace{1mm}14/litre$
and $1220$ litres of milk each week at $Rs\hspace{1mm}16/litre$
Now, if we assume the rate of milk as x-axis and Litres of milk as y-axis
Then, we will get coordinates of two points i.e. (14, 980) and (16, 1220)
Now, the relation between litres of milk and Rs/litres is given by equation
$(L-980)= \frac{1220-980}{16-14}(R-14)$
$(L-980)= \frac{240}{2}(R-14)$
$L-980= 120R-1680$
$L= 120R-700$
Now, at $Rs\hspace{1mm}17/litre$ he could sell
$L= 120\times 17-700= 2040-700= 1340$
He could sell 1340 litres of milk each week at $Rs\hspace{1mm}17/litre$

Question 17: $P(a,b)$ is the mid-point of a line segment between axes. Show that equation of the line is $\frac{x}{a}+\frac{y}{b}=2$.

Answer:

Now, let coordinates of point A is (0 , y) and of point B is (x , 0)
The,
$\frac{x+0}{2}= a \ and \ \frac{0+y}{2}= b$
$x= 2a \ and \ y = 2b$
Therefore, the coordinates of point A is (0 , 2b) and of point B is (2a , 0)
Now, slope of line passing through points (0,2b) and (2a,0) is
$m = \frac{0-2b}{2a-0} = \frac{-2b}{2a}= \frac{-b}{a}$
Now, equation of line passing through point (2a,0) and with slope $\frac{-b}{a}$ is
$(y-0)= \frac{-b}{a}(x-2a)$
$\frac{y}{b}= - \frac{x}{a}+2$
$\frac{x}{a}+\frac{y}{b}= 2$
Hence proved

Question 18: Point $R(h,k)$ divides a line segment between the axes in the ratio $1:2$ . Find equation of the line.

Answer:

Let the coordinates of Point A is (x,0) and of point B is (0,y)
It is given that point R(h , k) divides the line segment between the axes in the ratio $1:2$
Therefore,
R(h , k) $=\left ( \frac{1\times 0+2\times x}{1+2},\frac{1\times y+2\times 0}{1+2} \right )=\left ( \frac{2x}{3},\frac{y}{3} \right )$
$h = \frac{2x}{3} \ \ and \ \ k = \frac{y}{3}$
$x = \frac{3h}{2} \ \ and \ \ y = 3k$
Therefore, coordinates of point A is $\left ( \frac{3h}{2},0 \right )$ and of point B is $(0,3k)$
Now, slope of line passing through points $\left ( \frac{3h}{2},0 \right )$ and $(0,3k)$ is
$m = \frac{3k-0}{0-\frac{3h}{2}}= \frac{2k}{-h}$
Now, equation of line passing through point $(0,3k)$ and with slope $-\frac{2k}{h}$ is
$(y-3k)=-\frac{2k}{h}(x-0)$
$h(y-3k)=-2k(x)$
$yh-3kh=-2kx$
$2kx+yh=3kh$
Therefore, the equation of line is $2kx+yh=3kh$

Question 19: By using the concept of equation of a line, prove that the three points $(3,0),(-2,-2)$ and $(8,2)$ are collinear.

Answer:

Points are collinear means they lies on same line
Now, given points are $A(3,0),B(-2,-2)$ and $C(8,2)$
Equation of line passing through point A and B is
$(y-0)=\frac{0+2}{3+2}(x-3)$
$y=\frac{2}{5}(x-3)\Rightarrow 5y= 2(x-3)$
$2x-5y=6$
Therefore, the equation of line passing through A and B is $2x-5y=6$

Now, Equation of line passing through point B and C is
$(y-2)=\frac{2+2}{8+2}(x-8)$
$(y-2)=\frac{4}{10}(x-8)$
$(y-2)=\frac{2}{5}(x-8) \Rightarrow 5(y-2)=2(x-8)$
$5y-10=2x-16$
$2x-5y=6$
Therefore, Equation of line passing through point B and C is $2x-5y=6$
When can clearly see that Equation of line passing through point A nd B and through B and C is the same
By this we can say that points $A(3,0),B(-2,-2)$ and $C(8,2)$ are collinear points