Question

The acute angle bisector between the lines $3x-4y-5=0;5x+12y-26=0$ is

• A. $7x-56y+32=0$
• B. $9x-3y+13=0$
• C. $14x-112y+65=0$
• D. $7x-13y+9=0$

Answer 1

Answer: (C)

$14x-112y+65=0$

Locus of angle bisector:

Perpendicular distance of point $P(x_1,y_1)$ from both lines is equal,

$\Rightarrow \left|\frac{3x_1-4y_1-5}{5}\right|=\left|\frac{5x_1+12y_1-26}{13}\right|$

So, for acute angle bisector, we have to consider $+$ve sign

then locus of $p$ known as angle bisector of this two lines

$\because a_1{a}_2-b_1{b}_2<0$

$39x_1-52y_1-65=25x_1+60y_1-130$

$\Rightarrow 14x_1-112y_1+65=0$

The acute angled bisector is $14x-112y+65=0$