Question #10e39

Oct 15, 2017

$y = - \frac{1}{2} x + 3$

Explanation:

First, find the midpoint of the line segment.

$\left(\frac{8 + 4}{2} , \frac{4 + \left(- 4\right)}{2}\right)$
$= \left(6 , 0\right)$

Then, find the gradient of the line segment $\textcolor{p u r p \le}{m}$.
$\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

$= \frac{- 4 - 4}{4 - 8}$

$= \frac{- 8}{-} 4$

$= 2$

So the gradient of the bisector is $\frac{- 1}{\textcolor{p u r p \le}{m}} = = \frac{- 1}{2}$

Using the Cartesian equation $\Rightarrow y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 0 = - \frac{1}{2} \left(x - 6\right)$
$y = - \frac{1}{2} x + 3$