Examples (Revised) - Chapter 9 Straight Line class 11 ncert solutions Maths - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Chapter 9: Straight Line - NCERT Solutions for Class 11 Maths | Detailed Explanations

Example 1

Find the slope of the lines:
(a) Passing through the points $(3,-2)$ and $(-1,4)$,
(b) Passing through the points $(3,-2)$ and $(7,-2)$,
(c) Passing through the points $(3,-2)$ and $(3,4)$,
(d) Making inclination of $60^{\circ}$ with the positive direction of $x$-axis.

Solution

(a) The slope of the line through $(3,-2)$ and $(-1,4)$ is
$
m=\frac{4-(-2)}{-1-3}=\frac{6}{-4}=-\frac{3}{2} .
$
(b) The slope of the line through the points $(3,-2)$ and $(7,-2)$ is
$
m=\frac{-2-(-2)}{7-3}=\frac{0}{4}=0 \text {. }
$
(c) The slope of the line through the points $(3,-2)$ and $(3,4)$ is

$m=\frac{4-(-2)}{3-3}=\frac{6}{0}$, which is not defined.
(d) Here inclination of the line $\alpha=60^{\circ}$. Therefore, slope of the line is $m=\tan 60^{\circ}=\sqrt{3}$.

Example 2

If the angle between two lines is $\frac{\pi}{4}$ and slope of one of the lines is $\frac{1}{2}$, find the slope of the other line.
Solution

We know that the acute angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by
$
\tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|
$

Let $m_1=\frac{1}{2}, m_2=m$ and $\theta=\frac{\pi}{4}$.
Now, putting these values in (1), we get
$
\tan \frac{\pi}{4}=\left|\frac{m-\frac{1}{2}}{1+\frac{1}{2} m}\right| \text { or } 1=\left|\frac{m-\frac{1}{2}}{1+\frac{1}{2} m}\right|,
$
which gives
$
\frac{m-\frac{1}{2}}{1+\frac{1}{2} m}=1 \text { or } \frac{m-\frac{1}{2}}{1+\frac{1}{2} m}=-1 \text {. }
$

Therefore $m=3$ or $m=-\frac{1}{3}$. Hence, slope of the other line is 3 or $-\frac{1}{3}$. Fig 9.7 explains the reason of two answers.

Example 3

Line through the points $(-2,6)$ and $(4,8)$ is perpendicular to the line through the points $(8,12)$ and $(x, 24)$. Find the value of $x$.

Solution

Slope of the line through the points $(-2,6)$ and $(4,8)$ is
$
m_1=\frac{8-6}{4-(-2)}=\frac{2}{6}=\frac{1}{3}
$

Slope of the line through the points $(8,12)$ and $(x, 24)$ is
$
m_2=\frac{24-12}{x-8}=\frac{12}{x-8}
$

Since two lines are perpendicular, $m_1 m_2=-1$, which gives
$
\frac{1}{3} \times \frac{12}{x-8}=-1 \text { or } x=4 \text {. }
$

Example 4

Find the equations of the lines parallel to axes and passing through $(-2,3)$.

Solution

Position of the lines is shown in the Fig 9.9. The $y$-coordinate of every point on the line parallel to $x$-axis is 3 , therefore, equation of the line parallel to $x$-axis and passing through $(-2,3)$ is $y=3$. Similarly, equation of the line parallel to $y$-axis and passing through $(-2,3)$ is $x=-2$.

Example 5

Find the equation of the line through $(-2,3)$ with slope -4 .
Solution

Here $m=-4$ and given point $\left(x_0, y_0\right)$ is $(-2,3)$.
By slope-intercept form formula
(1) above, equation of the given line is
$y-3=-4(x+2)$ or $4 x+y+5=0$, which is the required equation.

Example 6

Write the equation of the line through the points $(1,-1)$ and $(3,5)$.
Solution

Here $x_1=1, y_1=-1, x_2=3$ and $y_2=5$. Using two-point form (2) above for the equation of the line, we have
$
y-(-1)=\frac{5-(-1)}{3-1}(x-1)
$
or $-3 x+y+4=0$, which is the required equation.

Example 7

Write the equation of the lines for which $\tan \theta=\frac{1}{2}$, where $\theta$ is the inclination of the line and (i) $y$-intercept is $-\frac{3}{2}$ (ii) $x$-intercept is 4 .

Solution

(i) Here, slope of the line is $m=\tan \theta=\frac{1}{2}$ and $y$ - intercept $c=-\frac{3}{2}$. Therefore, by slope-intercept form (3) above, the equation of the line is
$
y=\frac{1}{2} x-\frac{3}{2} \text { or } 2 y-x+3=0,
$
which is the required equation.
(ii) Here, we have $m=\tan \theta=\frac{1}{2}$ and $d=4$.

Therefore, by slope-intercept form (4) above, the equation of the line is
$
y=\frac{1}{2}(x-4) \text { or } 2 y-x+4=0,
$
which is the required equation.

Example 8

Find the equation of the line, which makes intercepts -3 and 2 on the $x$ - and $y$-axes respectively.
Solution

Here $a=-3$ and $b=2$. By intercept form (5) above, equation of the line is
$
\frac{x}{-3}+\frac{y}{2}=1 \quad \text { or } \quad 2 x-3 y+6=0 .
$

Any equation of the form $\mathrm{A} x+\mathrm{B} y+\mathrm{C}=0$, where $\mathrm{A}$ and $\mathrm{B}$ are not zero simultaneously is called general linear equation or general equation of a line.

Example 9

Find the distance of the point $(3,-5)$ from the line $3 x-4 y-26=0$.
Solution

Given line is $3 x-4 y-26=0$
Comparing (1) with general equation of line $\mathrm{A} x+\mathrm{B} y+\mathrm{C}=0$, we get
$
\mathrm{A}=3, \mathrm{~B}=-4 \text { and } \mathrm{C}=-26 \text {. }
$

Given point is $\left(x_1, y_1\right)=(3,-5)$. The distance of the given point from given line is
$
d=\frac{\left|\mathrm{A} x_1+\mathrm{B} y_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}=\frac{|3.3+(-4)(-5)-26|}{\sqrt{3^2+(-4)^2}}=\frac{3}{5} \text {. }
$

Example 10

Find the distance between the parallel lines $3 x-4 y+7=0$ and
$
3 x-4 y+5=0
$

Solution

Here $\mathrm{A}=3, \mathrm{~B}=-4, \mathrm{C}_1=7$ and $\mathrm{C}_2=5$. Therefore, the required distance is
$
d=\frac{|7-5|}{\sqrt{3^2+(-4)^2}}=\frac{2}{5} .
$