# A line segment is bisected by a line with the equation  4 y + x = 8 . If one end of the line segment is at (5 ,6 ), where is the other end?

Jul 25, 2016

Any point on the line $\textcolor{red}{4 y + x = - 17}$ together with $\left(5 , 6\right)$ will provide end points for a line segment bisected by $4 y + x = 8$

#### Explanation:

A vertical line through color(green)(""(5,6)) will intersect $\textcolor{p u r p \le}{4 y + x = 8}$
at color(purple)(""(5,3/4))

color(purple)(""(5,3/4)) is $5 \frac{1}{4}$ units below color(green)(""(5,6)

If color(purple)(""(5,3/4)) is the bisect point of the vertical line,
the second end point must be $5 \frac{1}{4}$ units lower than color(purple)(""(5,3/4))
That is the second end point must be at color(red)(""(5,-5 1/2))

The given line $\textcolor{p u r p \le}{4 y + x = 8}$ bisects the vertical line segment between color(green)(""(5,6)) and color(red)(""(5,-5 1/2))

Furthermore (based on similar triangles) any point on a line parallel to the bisector line $\textcolor{p u r p \le}{4 y + x = 8}$ will also be bisected by $\textcolor{p u r p \le}{4 y + x = 8}$

$\textcolor{p u r p \le}{4 y + x = 8}$ has a slope of $\left(- \frac{1}{4}\right)$
So all lines parallel to $\textcolor{p u r p \le}{4 y + x = 8}$ will have a slope of $\left(- \frac{1}{4}\right)$

The equation for a line passing through color(red)(""(5,-5 1/2)) with a slope of $\left(- \frac{1}{4}\right)$ can be expressed in slope-point form as
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{y - \left(- 5 \frac{1}{2}\right) = - \frac{1}{4} \left(x - 5\right)}$
or after simplifying, as
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{4 y + x = - 17}$