Miscellaneous Exercise (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 11

If the lines $2 x+y-3=0,5 x+k y-3=0$ and $3 x-y-2=0$ are concurrent, find the value of $k$.
Solution

Three lines are said to be concurrent, if they pass through a common point, i.e., point of intersection of any two lines lies on the third line. Here given lines are
$
\begin{aligned}
& 2 x+y-3=0 \\
& 5 x+k y-3=0 \\
& 3 x-y-2=0
\end{aligned}
$

Solving (1) and (3) by cross-multiplication method, we get
$
\frac{x}{-2-3}=\frac{y}{-9+4}=\frac{1}{-2-3} \quad \text { or } \quad x=1, y=1 .
$

Therefore, the point of intersection of two lines is $(1,1)$. Since above three lines are concurrent, the point $(1,1)$ will satisfy equation (2) so that
$
5.1+k .1-3=0 \text { or } k=-2 \text {. }
$

Example 12

Find the distance of the line $4 x-y=0$ from the point $\mathrm{P}(4,1)$ measured along the line making an angle of $135^{\circ}$ with the positive $x$-axis.
Solution

Given line is $4 x-y=0$ In order to find the distance of the line (1) from the point $\mathrm{P}(4,1)$ along another line, we have to find the point of intersection of both the lines. For this purpose, we will first find the equation of the second line (Fig 9.16). Slope of second line is $\tan 135^{\circ}=-1$. Equation of the line with slope -1 through the point $\mathrm{P}(4,1)$ is

$
y-1=-1(x-4) \text { or } x+y-5=0
$

Solving (1) and (2), we get $x=1$ and $y=4$ so that point of intersection of the two lines is $\mathrm{Q}(1,4)$. Now, distance of line (1) from the point $\mathrm{P}(4,1)$ along the line (2)
$
\begin{aligned}
& =\text { the distance between the points } \mathrm{P}(4,1) \text { and } \mathrm{Q}(1,4) \text {. } \\
& =\sqrt{(1-4)^2+(4-1)^2}=3 \sqrt{2} \text { units. }
\end{aligned}
$

Example 13

Assuming that straight lines work as the plane mirror for a point, find the image of the point $(1,2)$ in the line $x-3 y+4=0$.
Solution

Let $\mathrm{Q}(h, k)$ is the image of the point $\mathrm{P}(1,2)$ in the line
$
x-3 y+4=0
$

Therefore, the line (1) is the perpendicular bisector of line segment PQ (Fig 9.17).

Hence Slope of line $\mathrm{PQ}=\frac{-1}{\text { Slope of line } x-3 y+4=0}$,
so that $\frac{k-2}{h-1}=\frac{-1}{\frac{1}{3}}$ or $3 h+k=5$
and the mid-point of $\mathrm{PQ}$, i.e., point $\left(\frac{h+1}{2}, \frac{k+2}{2}\right)$ will satisfy the equation (1) so that
$
\frac{h+1}{2}-3\left(\frac{k+2}{2}\right)+4=0 \text { or } h-3 k=-3
$

Solving (2) and (3), we get $h=\frac{6}{5}$ and $k=\frac{7}{5}$.
Hence, the image of the point $(1,2)$ in the line (1) is $\left(\frac{6}{5}, \frac{7}{5}\right)$.
Example 14

Show that the area of the triangle formed by the lines $y=m_1 x+c_1, y=m_2 x+c_2$ and $x=0$ is $\frac{\left(c_1-c_2\right)^2}{2\left|m_1-m_2\right|}$.
Solution

Given lines are

$
x=\frac{\left(c_2-c_1\right)}{\left(m_1-m_2\right)} \text { and } y=\frac{\left(m_1 c_2-m_2 c_1\right)}{\left(m_1-m_2\right)}
$

Therefore, third vertex of the triangle is $\mathrm{R}\left(\frac{\left(c_2-c_1\right)}{\left(m_1-m_2\right)}, \frac{\left(m_1 c_2-m_2 c_1\right)}{\left(m_1-m_2\right)}\right)$.
Now, the area of the triangle is
$
=\frac{1}{2}\left|0\left(\frac{m_1 c_2-m_2 c_1}{m_1-m_2}-c_2\right)+\frac{c_2-c_1}{m_1-m_2}\left(c_2-c_1\right)+0\left(c_1-\frac{m_1 c_2-m_2 c_1}{m_1-m_2}\right)\right|=\frac{\left(c_2-c_1\right)^2}{2\left|m_1-m_2\right|}
$

Example 15

A line is such that its segment between the lines $5 x-y+4=0$ and $3 x+4 y-4=0$ is bisected at the point $(1,5)$. Obtain its equation.
Solution

Given lines are
$
\begin{aligned}
& 5 x-y+4=0 \\
& 3 x+4 y-4=0
\end{aligned}
$

Let the required line intersects the lines (1) and (2) at the points, $\left(\alpha_1, \beta_1\right)$ and $\left(\alpha_2, \beta_2\right)$, respectively
(Fig 9.19). Therefore
$
\begin{aligned}
& 5 \alpha_1-\beta_1+4=0 \text { and } \\
& 3 \alpha_2+4 \beta_2-4=0
\end{aligned}
$
or $\quad \beta_1=5 \alpha_1+4$ and $\beta_2=\frac{4-3 \alpha_2}{4}$.
We are given that the mid point of the segment of

the required line between $\left(\alpha_1, \beta_1\right)$ and $\left(\alpha_2, \beta_2\right)$ is $(1,5)$. Therefore
$
\frac{\alpha_1+\alpha_2}{2}=1 \text { and } \frac{\beta_1+\beta_2}{2}=5 \text {, }
$
or
$
\begin{aligned}
& \alpha_1+\alpha_2=2 \text { and } \frac{5 \alpha_1+4+\frac{4-3 \alpha_2}{4}}{2}=5, \\
& \text { or } \alpha_1+\alpha_2=2 \text { and } 20 \alpha_1-3 \alpha_2=20
\end{aligned}
$

Solving equations in (3) for $\alpha_1$ and $\alpha_2$, we get

$
\alpha_1=\frac{26}{23} \text { and } \alpha_2=\frac{20}{23} \text { and hence, } \beta_1=5 \cdot \frac{26}{23}+4=\frac{222}{23} .
$

Equation of the required line passing through $(1,5)$ and $\left(\alpha_1, \beta_1\right)$ is
$
y-5=\frac{\beta_1-5}{\alpha_1-1}(x-1) \text { or } y-5=\frac{\frac{222}{23}-5}{\frac{26}{23}-1}(x-1)
$
or
$
107 x-3 y-92=0,
$
which is the equation of required line.
Example 16

Show that the path of a moving point such that its distances from two lines $3 x-2 y=5$ and $3 x+2 y=5$ are equal is a straight line.
Solution

Given lines are
$
\begin{aligned}
& 3 x-2 y=5 \\
& 3 x+2 y=5
\end{aligned}
$
and
Let $(h, k)$ is any point, whose distances from the lines (1) and (2) are equal. Therefore
$
\frac{|3 h-2 k-5|}{\sqrt{9+4}}=\frac{|3 h+2 k-5|}{\sqrt{9+4}} \text { or }|3 h-2 k-5|=|3 h+2 k-5| \text {, }
$
which gives $3 h-2 k-5=3 h+2 k-5$ or $-(3 h-2 k-5)=3 h+2 k-5$.
Solving these two relations we get $k=0$ or $h=\frac{5}{3}$. Thus, the point $(h, k)$ satisfies the equations $y=0$ or $x=\frac{5}{3}$, which represent straight lines. Hence, path of the point equidistant from the lines (1) and (2) is a straight line.