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
$$\begin{gathered} (y-3)=\frac{1}{2}(x-(-4)) \\ 2y-6=x+4 \\ x-2y+10=0 \end{gathered}$$
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
$$\begin{gathered} (y-0)=m(x-0) \\ y=mx \end{gathered}$$
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
$m=\tan {75}^{\circ }(\because \theta ={75}^{\circ }given)$
$m=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

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
$$\begin{gathered} (y-0)=-2(x-(-3)) \\ y=-2x-6 \\ 2x+y+6=0 \end{gathered}$$
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
$$\begin{gathered} m=\tan \theta \\ m=\tan {30}^{\circ }(\because \theta ={30}^{\circ }given) \\ m=\frac{1}{\sqrt{3}} \end{gathered}$$
Now, the equation of the line passing through the point (0,2) and with slope $\frac{1}{\sqrt{3}}$ is
$$\begin{gathered} (y-2)=\frac{1}{\sqrt{3}}(x-0) \\ \sqrt{3}(y-2)=x \\ x-\sqrt{3}y+2\sqrt{3}=0 \end{gathered}$$
Therefore, the equation of the line is $x-\sqrt{3}y+2\sqrt{3}=0$

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)
$$\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{-4-1}{2+1}=\frac{-5}{3} \end{gathered}$$
Now, equation of line passing through point (-1,1) and with slope $\frac{-5}{3}$ is
$$\begin{gathered} (y-1)=\frac{-5}{3}(x-(-1)) \\ 3(y-1)=-5(x+1) \\ 3y-3 \end{gathered}$$

$$\begin{gathered} =-5x-5 \\ 5x+3y+2=0 \end{gathered}$$

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

Answer:

The vertices of $\mathrm{\Delta }PQR$ 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 ) = $(\frac{2-2}{2},\frac{1+3}{2})=(0,2)$
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
$$\begin{gathered} (y-2)=\frac{3}{4}(x-0) \\ 4(y-2)=3x \\ 4y-8 \end{gathered}$$

$$\begin{gathered} =3x \\ 3x-4y+8=0 \end{gathered}$$
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)
$m^{\prime }=\frac{6-5}{-3-2}=\frac{1}{-5}$
Now this line is perpendicular to line passing through point (-3,5)
Therefore,
$m=-\frac{1}{m^{\prime }}=-\frac{1}{\frac{1}{-5}}=5$

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
$$\begin{gathered} (y-5)=5(x-(-3)) \\ (y-5)=5(x+3) \\ y-5 \end{gathered}$$

$$\begin{gathered} =5x+15 \\ 5x-y+20=0 \end{gathered}$$
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
$(\frac{n(1)+1(2)}{1+n},\frac{n.(0)+1.(3)}{1+n})=(\frac{n+2}{1+n},\frac{3}{1+n})$
Let the slope of the perpendicular line is m
And Slope of line segment joining the points $(1,0)$ and $(2,3)$ is
$m^{\prime }=\frac{3-0}{2-1}=3$
Now, slope of perpendicular line is
$m=-\frac{1}{m^{\prime }}=-\frac{1}{3}$
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 $(\frac{n+2}{1+n},\frac{3}{1+n})$ and with slope $-\frac{1}{3}$ is
$$\begin{gathered} (y-\frac{3}{1+n})=-\frac{1}{3}(x-(\frac{n+2}{1+n})) \\ 3y(1+n)-9 \end{gathered}$$

$$\begin{gathered} =-x(1+n)+n+2 \\ x(1+n)+3y(1+n)=n+11 \end{gathered}$$
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
$$\begin{gathered} \frac{x}{a}+\frac{y}{a}=1 \\ x+y=a \end{gathered}$$
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,
$$\begin{gathered} \frac{x}{a}+\frac{y}{9-a}=1 \\ x(9-a)+ay=a(9-a) \\ 9x-ax+ay \end{gathered}$$

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

$$\begin{gathered} =0 \\ {a}^2-6a-3a+18 \end{gathered}$$

$$\begin{gathered} =0 \\ (a-6)(a-3)=0 \\ a=6ora=3 \end{gathered}$$

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

case (ii) a = 3 , b = 6
$$\begin{gathered} \frac{x}{3}+\frac{y}{6}=1 \\ 2x+y=6 \end{gathered}$$
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
$$\begin{gathered} m=\tan \theta \\ m=\tan \frac{2\pi }{3}=-\sqrt{3} \end{gathered}$$
Now, equation of line passing through point (0 , 2) and with slope $-\sqrt{3}$ is
$$\begin{gathered} (y-2)=-\sqrt{3}(x-0) \\ \sqrt{3}x+y-2=0 \end{gathered}$$
Therefore, equation of line is $\sqrt{3}x+y-2=0$ -(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 $-\sqrt{3}$ is
$$\begin{gathered} (y-(-2))=-\sqrt{3}(x-0) \\ \sqrt{3}x+y+2=0 \end{gathered}$$
Therefore, equation of line is $\sqrt{3}x+y+2=0$

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
$m^{\prime }=\frac{9-0}{-2-0}=\frac{9}{-2}$
Now, the slope of the line is
$m=-\frac{1}{m^{\prime }}=\frac{2}{9}$
Now, the equation of line passes through the point (-2, 9) and with slope $\frac{2}{9}$ is
$$\begin{gathered} (y-9)=\frac{2}{9}(x-(-2)) \\ 9(y-9) \end{gathered}$$

$$\begin{gathered} =2(x+2) \\ 2x-9y+85=0 \end{gathered}$$
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 $Rs14/litre$ and $1220$ litres of milk each week at $Rs16/litre$ . Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at $Rs17/litre$

Answer:

It is given that the owner of a milk store sell
980 litres milk each week at $Rs14/litre$
and $1220$ litres of milk each week at $Rs16/litre$
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 $Rs17/litre$ he could sell
$L=120\times 17-700=2040-700=1340$
He could sell 1340 litres of milk each week at $Rs17/litre$

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,
$\frac{x+0}{2}=aand\frac{0+y}{2}=b$
$x=2aandy=2b$
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) $=(\frac{1\times 0+2\times x}{1+2},\frac{1\times y+2\times 0}{1+2})=(\frac{2x}{3},\frac{y}{3})$
$h=\frac{2x}{3}andk=\frac{y}{3}$
$x=\frac{3h}{2}andy=3k$
Therefore, coordinates of point A is $(\frac{3h}{2},0)$ and of point B is $(0,3k)$
Now, slope of line passing through points $(\frac{3h}{2},0)$ 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