Bisector

• Feb 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.

• Feb 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 $\displaystyle y = mx + c$ form...

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

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

So in the first case:

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

In the second case:

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

The angle between them is

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

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

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