Find the equation of the bisector of the acute angle formed by the lines 3x-y-5=0 and 3x+4y-12=0.
please help me and show the solution.. thanks!
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 $\displaystyle c$.