Question

The line $x+3y-2=0$ bisects the angle between a pair of straight lines of which one has equation $x-7y+5=0$, then find equation of other line.

Answer 1

Let,

$L_1$ denotes $x+3y-2=0$

$L_2$ denotes $x-+7y+5=0$ and

$L_3$ denotes the line whose equation is to be found out.

As the line $L_1$ is the angle bisector.

This means that the angle between the line $L_1$ and $L_2$ will be equal to the the angle between the line $L_1$ and $L_3$

Let the angle b/w $L_1$ and $L_2$ be $\theta$ and the slope of line $L_1$, $L_2$.and $L_3$ be $m_1$, $m_2$ and $m_3$ respectively.

From the equation of line, we can see that $m_1=\frac{-1}{3}$ and $m_2=\frac{1}{7}$

$\Rightarrow \tan \theta =\left|\frac{\frac{1}{7}-\frac{-1}{3}}{1+\frac{1}{7}\times \frac{-1}{3}}\right|$

$\Rightarrow \tan \theta =\frac{1}{2}$

Now, the angle b/w $L_1$ and $L_3$ should also be $\theta$

$\Rightarrow \frac{1}{2}=\left|\frac{\frac{-1}{3}-m_3}{1+m_3\times \frac{-1}{3}}\right|$

On solving we get $m_3=-1$ or $\frac{1}{7}$

Now, if $m_3=\frac{1}{7}$ the $L_3$ would become same as $L_2$

This means that $m_3=-1$

Now, we have to find $P$ i.e. the point of intersection of the lines $L_1$ and $L_2$ in order to apply one point formula to obtain the equation of $L_3$ as $L_3$ will also pass through point $P$

On solving $L_1$ and $L_2$ we find that the point $P=(\frac{-1}{10},\frac{7}{10})$

Applying the one point formula with $P$ and $m_3$, we get

$\Rightarrow (y-\frac{7}{10})=(-1)(x-\left(\frac{-1}{10}\right))$

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

Hence, the equation of the required line $L_3$ is $5x+5y-3=0$