# A line segment is bisected by a line with the equation  - 6 y + 2 x = 3 . If one end of the line segment is at ( 4 , 8 ), where is the other end?

Jun 21, 2016

Any point on the line $3 y - x = - 27$

#### Explanation:

Consider a horizontal line segment from $\left(4 , 8\right)$ bisected by $- 6 y + 2 x = 3$
The equation of the horizontal line segment from $\left(4 , 8\right)$ is
$\textcolor{w h i t e}{\text{XXX}} y = 8$

Noting that $- 6 y + 2 x = 3 \Rightarrow x = \frac{6 y + 3}{2}$

The intersection of this horizontal line segment with the given bisector line will occur at
$\textcolor{w h i t e}{\text{XXX}} \left(\frac{6 \left(8\right) + 3}{2} , 8\right) = \left(\frac{51}{2} , 8\right)$

Continuing to travel horizontally a point twice as far away from $\left(4 , 8\right)$ as $\left(\frac{51}{2} , 8\right)$
will be at $\left(51 , 8\right)$

That is $\left(51 , 8\right)$ is one possible endpoint for a line segment form $\left(4 , 8\right)$ bisected by $- 6 y + 2 x = 3$ would be $\left(51 , 8\right)$

For any point which could be such an endpoint, a line through this point parallel to the bisecting line will provide all possible bisected line segment endpoints.

$- 6 y + 2 x = 3$ (and all lines parallel to it) has a slope of $\frac{1}{3}$

Using the previously determined possible endpoint $\left(51 , 8\right)$ and this slope of $\frac{1}{3}$
we can write the slope-point version:
$\textcolor{w h i t e}{\text{XXX}} y - 8 = \frac{1}{3} \left(x - 51\right)$

$\textcolor{w h i t e}{\text{XXX}} 3 y - 24 = x - 51$

$\textcolor{w h i t e}{\text{XXX}} 3 y - x = - 27$

Any solution to the equation $3 y - x = - 27$ will provide a valid endpoint for a line segment from $\left(4 , 8\right)$ which will be bisected by $6 y - 2 x = 3$

Here is a graph of the point and the two lines in question:
graph{(-6y+2x-3)(3y-x+27)((x-4)^2+(y-8)^2-0.02)=0 [-25.65, 25.64, -12.83, 12.81]}