Question

Find the equations of bisectors of the angle between the lines $4x+3y=7$ and $24x+7y-31=0$. Also find which of them is the bisector of the acute angle.

• A. $2x+y-3=0$
• B. $x-2y+1=0$
• C. $x-2y-1=0$
• D. $2x-y-3=0$

Answer 1

The correct option is A $2x+y-3=0$

Given, $4x+3y=7$ and $24x+7y-31=0$

Here, $A_1=4$, $B_1=3$, $C_1=7$, $A_2=24$, $B_2=7$, $C_2=-31$

Equation of angle bisector of lines is

$\left|\frac{A_{1}x+B_{1}y+C_1}{\sqrt{A_1^2+B_1^2}}\right|$ $=\pm \left|\frac{A_{2}x+B_{2}y+C_2}{\sqrt{A_2^2+B_2^2}}\right|$

$4x+3y-7=0$ and $24x+7y-31=0$ is

$\left|\frac{4x+3y-7}{\sqrt{4^2+3^2}}\right|=\pm \left|\frac{24x+7y-31}{\sqrt{{\left(24\right)}^2+7^2}}\right|$

$\Rightarrow \left|\frac{4x+3y-7}{5}\right|=\pm \left|\frac{24x+7y-31}{25}\right|$

As $-7$ and $-31$ are of same sign and $a_1{a}_2+b_1{b}_2=4.24+3.7>0$

Then equation containing acute angle is
$\left|\frac{4x+3y-7}{5}\right|=-\left|\frac{24x+7y-31}{25}\right|$

$\Rightarrow 20x+15y-35=-24x-7y+31$

$\Rightarrow 44x+22y-66=0\Rightarrow 2x+y-3=0$