# Analytic Geometry Q11

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• Mar 4th 2009, 11:16 AM
looi76
Analytic Geometry Q11
Question:
Find the equation of the line that bisects the acute angles determined by the given lines: $\displaystyle x-6y-12=0$ and $\displaystyle 4x-6y-9=0$

Attempt:

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

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

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

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

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

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

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

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

Is my answer correct?
• Mar 4th 2009, 10:33 PM
earboth
Quote:

Originally Posted by looi76
...

Is my answer correct?

Yes.

I've attached a graph of the three lines.