Line Equally Inclined with Two Lines

In this article, we will cover the concept of Line Equally Inclined with two lines. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of ten questions have been asked on JEE MAINS( 2013 to 2023) from this topic.

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This Story also Contains

1. What is Bisector?
2. Line Equally Inclined with two lines
3. Derivation of Line Equally Inclined with two lines
4. Solved Examples Based on Line Equally Inclined with two lines

What is Bisector?

The bisector is the locus of a point that moves in the plane of lines L₁ and L₂ such that lengths of perpendiculars drawn from it to the two given lines(L₁ and L₂) are equal.

Line Equally Inclined with two lines

A line equally inclined with two lines means the line which is the angle bisector of the angle made by two lines.

If the two lines with slope $m_1$ and $m_2$ are equally inclined to a line having slope $m$, then
$
\left(\frac{\mathbf{m}_1-\mathbf{m}}{1+\mathbf{m}_1 \mathbf{m}}\right)=-\left(\frac{\mathbf{m}_2-\mathbf{m}}{1+\mathbf{m}_2 \mathbf{m}}\right)
$

Derivation of Line Equally Inclined with two lines

Two lines with slopes $m_1$ and $m_2$ intersect at point A

As from the fig

$
\begin{aligned}
& \angle \mathrm{PAQ}=\angle \mathrm{QAR}=\angle \theta \\
& \tan (\angle \mathrm{PAQ})=\frac{\mathrm{m}_1-\mathrm{m}}{1+\mathrm{m}_1 \mathrm{~m}}=\tan \theta \\
& \text { and } \\
& \tan (\angle \mathrm{QAR})=\frac{\mathrm{m}-\mathrm{m}_2}{1+\mathrm{mm}_2}=\tan \theta
\end{aligned}
$

Hence,
$
\left(\frac{\mathbf{m}_1-\mathbf{m}}{1+\mathbf{m}_1 \mathbf{m}}\right)=-\left(\frac{\mathbf{m}_2-\mathbf{m}}{1+\mathbf{m}_2 \mathbf{m}}\right)
$

Solved Examples Based on Line Equally Inclined with two lines

Example 1: Equations of line $\mathrm{L}_1$ is $\sqrt{3} y-x+1=0$ and Equation of line $\mathrm{L}_2$ is $y-\sqrt{3} x+1=0$ then find the equation of Line which is equally inclined with both line and passing through the intersection point
Solution: If the two lines with slope $m_1$ and $m_2$ are equally inclined to a line having slope $m$, then
$
\left(\frac{\mathbf{m}_1-\mathbf{m}}{1+\mathbf{m}_1 \mathbf{m}}\right)=-\left(\frac{\mathbf{m}_2-\mathbf{m}}{1+\mathbf{m}_2 \mathbf{m}}\right)
$

Two lines with slopes $m_1$ and $m_2$ intersect at point $A$

As from the figure

$
\begin{aligned}
& \angle \mathrm{PAQ}=\angle \mathrm{QAR}=\angle \theta \\
& \tan (\angle \mathrm{PAQ})=\frac{\mathrm{m}_1-\mathrm{m}}{1+\mathrm{m}_1 \mathrm{~m}}=\tan \theta \\
& \text { and } \\
& \tan (\angle \mathrm{QAR})=\frac{\mathrm{m}-\mathrm{m}_2}{1+\mathrm{mm}_2}=\tan \theta
\end{aligned}
$

Hence,
$
\begin{aligned}
& \left(\frac{\mathbf{m}_1-\mathbf{m}}{1+\mathbf{m}_1 \mathbf{m}}\right)=-\left(\frac{\mathbf{m}_2-\mathbf{m}}{1+\mathbf{m}_2 \mathbf{m}}\right) \\
& \quad \sqrt{3} y-x+1=0 \Rightarrow m_1=\frac{1}{\sqrt{3}} \\
& y-\sqrt{3} x+1=0 \Rightarrow m_2=\sqrt{3}^y y-\sqrt{3} x+1=0
\end{aligned}
$

If the two lines with slope $m_1$ and $m_2$ are equally inclined to a line having slope $m$, then $\left(\frac{m_1-m}{1+m_1 m}\right)=-\left(\frac{m_2-m}{1+m_2 m}\right)$
$
\begin{aligned}
& m= \pm 1 \\
& \quad\left(\frac{\frac{1}{\sqrt{3}}-\mathbf{m}}{1+\frac{1}{\sqrt{3}} \mathbf{m}}\right)=-\left(\frac{\sqrt{3}-\mathbf{m}}{1+\sqrt{3} \mathbf{m}}\right) \Rightarrow m^2=1 \\
& m
\end{aligned}
$

For Intersection point
$
\begin{aligned}
& \sqrt{3} y-x+1=0 \\
& y-\sqrt{3} x+1=0 \\
& x=\frac{\sqrt{3}-1}{2} \\
& y=\frac{3}{2}(1-\sqrt{3})
\end{aligned}
$

Equation of lines
$
\begin{aligned}
& y-\frac{3}{2}(1-\sqrt{3})=x-\frac{\sqrt{3}-1}{2} \\
& L_1: y=x+2-2 \sqrt{3} \\
& y-\frac{3}{2}(1-\sqrt{3})=-1\left(x-\frac{\sqrt{3}-1}{2}\right) \\
& L_2: y=-x+1-\sqrt{3}
\end{aligned}
$

Hence, the answer is $y=x+2-2 \sqrt{3}$

Example 2: A line is passing through (1,1) and $(2,1+\sqrt{3})$ and another line which is passing through (1,1) and $\left(2,1+\frac{1}{\sqrt{3}}\right)$ then find the equation of the line which is equally inclined with both lines.
Solution: If the two lines with slope $m_1$ and $m_2$ are equally inclined to a line having slope $m$, then
$
\left(\frac{\mathbf{m}_1-\mathbf{m}}{1+\mathbf{m}_1 \mathbf{m}}\right)=-\left(\frac{\mathbf{m}_2-\mathbf{m}}{1+\mathbf{m}_2 \mathbf{m}}\right)
$

Two lines with slopes $m_1$ and $m_2$ intersect at point $A$

As from the figure
$
\begin{aligned}
& \angle \mathrm{PAQ}=\angle \mathrm{QAR}=\angle \theta \\
& \tan (\angle \mathrm{PAQ})=\frac{\mathrm{m}_1-\mathrm{m}}{1+\mathrm{m}_1 \mathrm{~m}}=\tan \theta
\end{aligned}
$
and
$
\tan (\angle \mathrm{QAR})=\frac{\mathrm{m}-\mathrm{m}_2}{1+\mathrm{mm}_2}=\tan \theta
$

Hence,
$
\left(\frac{\mathbf{m}_1-\mathbf{m}}{1+\mathbf{m}_1 \mathbf{m}}\right)=-\left(\frac{\mathbf{m}_2-\mathbf{m}}{1+\mathbf{m}_2 \mathbf{m}}\right)
$

Slope of line $L_1=\frac{1+\sqrt{3}-1}{2-1}=\sqrt{3}=\tan 60^{\circ}$
Slope of line $L_2=\frac{1+\frac{1}{\sqrt{3}}-1}{2-1}=\frac{1}{\sqrt{3}}=\tan 30^{\circ}$
Slope of line which is equally inclined $=\tan 45^{\circ}=1$ and another line which is also equally inclined,slope is $=-1$ Equations of lines
$
\begin{aligned}
& L_1: y-1=x-1 \\
& \Rightarrow y-x=0 \\
& L_2: y-1=-1(x-1) \\
& \Rightarrow y+x-2=0
\end{aligned}
$

Hence, the answer is $y+x-2=0$

Example 3: Two equal sides of an isosceles triangle have the equations $7 x-y+3=0$ and $x+y=3$ and its third side passes through the point ( $-2,-1$ ). Then the equation of the third side.
Solution: If the two lines with slope $m_1$ and $m_2$ are equally inclined to a line having slope $m$, then
Two lines with slopes $m_1$ and $m_2$ intersect at point $A$

Let Equation of $A B$ is $7 x-y+3=0$ and equation of $A C x+y=3$
Slope of $\mathrm{AB}=7$ and Slope of $\mathrm{AC}=-1$
Let the equation of third side $B C=y=m x+c$
The angle between $A B$ and $B C=$ Angle between $A C$ and $B C$
$
\begin{aligned}
& \left|\frac{7-m}{1+7 m}\right|=\left|\frac{-1-m}{1-m}\right| \\
& -6 m^2-16 m+6=0 \\
& 3 m^2+8 m-3=0 \\
& m=-3 \text { and } m=1 / 3
\end{aligned}
$
equation of $B C$ is
$
(y-(-1))=-3(x-(-2)) \Rightarrow 3 x+y+7=0
$
or
$
(y-(-1))=\frac{1}{3}(x-(-2)) \Rightarrow x-3 y=1
$

Hence, the answer is $x-3 y=1$

Example 4: If $5 x^2+\lambda y^2=20$ represents a rectangular hyperbola, then $\lambda$ equal
Solution: The general equation of the second degree represents a rectangular hyperbola if $\Delta \neq 0, h^2>a b$ and coefficient of $x^2+$ coefficient of $y^2=0$.

The given equation represents a rectangular hyperbola if $\lambda+5=0$ i.e., $\lambda=-5$.
Hence, the answer is -5.

Example 5: A ray of light travelling along the line $2 \mathrm{x}-3 \mathrm{y}+5=0$ after striking a plane mirror lying along the line $\mathrm{x}+\mathrm{y}=2$ gets reflected. Find the equation of the straight line containing the reflected ray

Solution: The point of intersection of the lines $2 \mathrm{x}-3 \mathrm{y}+5=0$ and $\mathrm{x}+\mathrm{y}=2$ is $\left(\frac{1}{5}, \frac{9}{5}\right)$.
$\left(\frac{1}{5}, \frac{9}{5}\right)$ is the point of incidence.
The slope $m$ of the normal to the mirror (i.e. normal to the line $x+y=2$ ) is 1
Now the incident ray and reflected ray both are equally inclined to the normal and are on opposite sides of it.
The slope of the incident ray
$
\mathrm{m}_{\mathrm{I}}=\frac{2}{3}
$

Let the slope of the reflected ray be $=\mathrm{m}_2$
Then
$\frac{m_1-m}{1+m_1 \mathrm{~m}}=\frac{m-m_2}{1+m_2 m}$
$
\frac{\frac{2}{3}-1}{1+\frac{2}{3} \times 1}=\frac{1-m_2}{1+m_2 \times 1}
$

$\therefore \mathrm{m}_2=\frac{3}{2}, \quad \therefore$ the equation of the straight line containing the reflected ray is
$y-\frac{9}{5}=\frac{3}{2}\left(x-\frac{1}{5}\right)$
i.e. $\quad 3 x-2 y+3=0$

Hence, the answer is $3 x-2 y+3=0$