# Bisector

• February 19th 2010, 11:05 PM
lance
Bisector
Find the equation of the bisector of the acute angle formed by the lines 3x-y-5=0 and 3x+4y-12=0.

• February 19th 2010, 11:38 PM
Prove It
Quote:

Originally Posted by lance
Find the equation of the bisector of the acute angle formed by the lines 3x-y-5=0 and 3x+4y-12=0.

Write the equations of the lines in $y = mx + c$ form...

$y = 3x - 5$ and $y = -\frac{3}{4}x + 3$.

Note that the gradient of any line $m = \tan{\theta}$, where $\theta$ is the angle made with the positive $x$ axis.

So in the first case:

$\tan{\theta_1} = 3$ and so $\theta_1 = \arctan{3}$.

In the second case:

$\tan{\theta_2} = -\frac{3}{4}$, so $\theta_2 = \arctan{\left(-\frac{3}{4}\right)}$.

The angle between them is

$\theta = \frac{\theta_1 + \theta_2}{2}$.

Therefore the gradient of the line is $m = \tan{\theta}$.

Now, work out the point of intersection of the two lines, and you can substitute this point into $y = mx + c$ to find $c$.