# A line segment is bisected by a line with the equation  -7 y + x = 3 . If one end of the line segment is at (1 ,6 ), where is the other end?

May 2, 2016

The other end of the bisected line could be any point on
$\textcolor{w h i t e}{\text{XXX}} - 7 y + x = 47$

#### Explanation:

If $\textcolor{red}{- 7 y + x = 3}$ bisects a line segment between color(blue)(""(1,6)) and some other point color(green)(""(x',y'))
with an intersection point of $\textcolor{red}{- 7 y + x = 3}$ and this line segment at color(red)(""(barx,bary))

Then
$\textcolor{w h i t e}{\text{XXX}} \left(6 - \textcolor{red}{\overline{y}}\right) = \left(\textcolor{red}{\overline{y}} - \textcolor{g r e e n}{y '}\right)$
$\textcolor{w h i t e}{\text{XXXXXX}} \rightarrow \textcolor{g r e e n}{y '} = 2 \textcolor{red}{\overline{y}} - 6$
and
$\textcolor{w h i t e}{\text{XXX}} \left(1 - \textcolor{red}{\overline{x}}\right) = \left(\textcolor{red}{\overline{x}} - \textcolor{g r e e n}{x '}\right)$
$\textcolor{w h i t e}{\text{XXXXXX}} \rightarrow \textcolor{g r e e n}{x '} = 2 \textcolor{red}{\overline{x}} - 1$

Specifically, we could solve $\textcolor{red}{- 7 y + x = 3}$ for a couple arbitrary solution points:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{\left({\overline{x}}_{1} , {\overline{y}}_{1}\right) = \left(- 4 , - 1\right)}$
and
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{\left({\overline{x}}_{2} , {\overline{y}}_{2}\right) = \left(3 , 0\right)}$

and obtain end-of-line-segment values:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{\left(x {'}_{1} , y {'}_{1}\right) = \left(- 9 , - 8\right)}$
and
color(white)("XXX")color(green)((x'_2,y'_2)=(5,-6)

Using the slopes
$\textcolor{w h i t e}{\text{XXX}} \frac{y - \textcolor{g r e e n}{y {'}_{1}}}{x - \textcolor{g r e e n}{x {'}_{1}}} = \frac{\textcolor{g r e e n}{y {'}_{1} - y {'}_{2}}}{\textcolor{g r e e n}{x {'}_{1} - x {'}_{2}}}$

color(white)("XXX")(y-color(green)(""(-8)))/(x-color(green)(""(-9)))=(color(green)(""(-8))-color(green)(""(-6)))/(color(green)(""(-9))-color(green)(""(-5)))

$\textcolor{w h i t e}{\text{XXX}} \frac{y + 8}{x + 9} = \frac{1}{7}$

$\textcolor{w h i t e}{\text{XXX}} 7 y + 56 = x + 9$

$\textcolor{w h i t e}{\text{XXX}} 7 y - x = - 47$

or (paralleling the given form):
$\textcolor{w h i t e}{\text{XXX}} - 7 y + x = 47$