Analytic Geometry Q11

**looi76** · March 4th 2009, 11:16 AM

Question:
Find the equation of the line that bisects the acute angles determined by the given lines: $x-6y-12=0$ and $4x-6y-9=0$

Attempt:

$L_1 \rightarrow x-6y-12=0$

$d_1 = \frac{Ax+By+C}{\pm\sqrt{A^2+B^2}} = \frac{x'-6y'-12}{\sqrt{1^2+6^2}} = \frac{x'-6y'-12}{+\sqrt{37}}$

$L_2 \rightarrow 4x-6y-9=0$

$d_2 = \frac{Ax+By+C}{\pm\sqrt{A^2+B^2}} = \frac{4x'-6y'-9}{\sqrt{4^2+6^2}} = \frac{4x'-6y'-9}{-2\sqrt{13}}$

Since, the bisect line is in the middle of the lines then we use the formula $\boxed{d_1+d_2=0}$

$\left[\frac{x'-6y'-12}{+\sqrt{37}} + \frac{4x'-6y'-9}{-2\sqrt{13}} = 0\right]\sqrt{37}\times -2\sqrt{13}$

$-2\sqrt{13}x' +12\sqrt{13}y' +24\sqrt{13} + 4\sqrt{37}x' - 6\sqrt{37}y' -9\sqrt{37} = 0$

Answer:
$\boxed{(-2\sqrt{13} + 4\sqrt{37})x' + (12\sqrt{13} - 6\sqrt{37})y' + (24\sqrt{13} - 9\sqrt{37}) = 0}$

Is my answer correct?