# A line segment is bisected by a line with the equation  - 2 y - x = 1 . If one end of the line segment is at ( 8 , 3 ), where is the other end?

May 31, 2016

$- 2 y - x = 1$ bisects line segments between $\left(8 , 3\right)$ and any point on the line $- 2 y - x = 16$

#### Explanation:

Consider a horizontal line segment from $\left(8 , 3\right)$ extending to the left
with a total length twice that of the horizontal distance from $\left(8 , 3\right)$ to the line $- 2 y - x = 1$.
From the diagram below we can see that this horizontal line will intersect $- 2 y - x = 1$ at $\left(- 7 , 3\right)$
and will have its left-most end at $\left(- 22 , 3\right)$ Obviously this horizontal line is bisected by $- 2 y - x = 1$

Consider a line through this left-most point with the same slope as $- 2 y - x = 1$.
Using the slope-point form we can see that the equation of this line can be written as $- 2 y - x = 16$

Assertion
for any point $\left(x ' , y '\right)$ on this new line, the line segment between $\left(x ' , y '\right)$ and $\left(8 , 3\right)$ will be bisected by $- 2 y - x = 1$

This Assertion follows immediately from considering a modified version of the above diagram. Since $Q R | | S T$
color(white)("XXX")trianglePQR ~ trianglePST

$\textcolor{w h i t e}{\text{XXX}} \Rightarrow P R : R T = P Q : Q S = 1 : 1$

$\Rightarrow Q R$ bisects $P T$ for any point $T = \left(x ' , y '\right)$ on $- 2 y - x = 16$