Question

The equation of the bisector of the acute angle between the lines $3x-4y+7=0$ and $12x+5y-2=0$ is

• A. $11x+3y-9=0$
• B. $3x-11y+9=0$
• C. $11x-3y-9=0$
• D. $11x-3y+9=0$

Answer 1

The correct option is D $11x-3y+9=0$

$AS=SD=AD=2x$

$=2\times 10\sqrt{3}$
$=20\sqrt{3}m$

Length of string of each phone is $20\sqrt{3}m$

$\sqrt{3x}=20+\sqrt{400-x^2}$

$\sqrt{400-x^2}=\sqrt{3}x-20$

by solving

$40\sqrt{3}x=4x^2$

$\frac{x^2}{x}=\frac{40}{4}\sqrt{3}$

$x=10\sqrt{3}$

$3x-4y+7=0$

$12x+5y-2=0\Rightarrow -12x-5y+2=0$

$a{a}_1+b_1{b}_2=3(-12)-\left(4\right)(-5)<0$

hence the positive sign gives us the acute angle bisector which is

$3x-4y+7/5=-12x-5y+\frac{2}{13}=0$

This yields $99x-27y+81=0$

$11x-3y+9=0$