# A line segment is bisected by a line with the equation  -6 y + 3 x = 2 . If one end of the line segment is at ( 5 , 1 ), where is the other end?

Sep 12, 2016

Any point on the line $y = \frac{x}{2} + \frac{5}{6}$

#### Explanation:

$\textcolor{red}{- 6 y + 3 x = 2}$ can be rewritten as $\textcolor{red}{y = \frac{1}{2} x - \frac{1}{3}}$

When $x = 5$
$\textcolor{w h i t e}{\text{XXX")y=1/2x-1/3color(white)("XX")rarrcolor(white)("XX}} y = \textcolor{red}{\frac{13}{6}}$

The vertical distance from color(green)(""(5,1)) to the point color(red)(""(5,13/6)) on the line $\textcolor{red}{- 6 y + 3 x = 2}$ is
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{\frac{13}{6} - 1 = \frac{7}{6}}$

That is moving vertically the line $\textcolor{red}{- 6 y + 3 x = 2}$ is $\textcolor{b l u e}{\frac{7}{6}}$ above the point color(green)(""(5,1)).

If we continue moving upward another $\textcolor{b l u e}{\frac{7}{6}}$ units (for a total of $\frac{14}{6} = \frac{7}{3}$ units above color(green)(""(5,1)))
we will reach the point ""(5,10/3))
which will be the same distance above $\textcolor{red}{- 6 y + 3 x = 2}$ as color(green)(""(5,1)) is below it.

That is $\textcolor{red}{- 6 y + 3 x = 2}$ bisects the line segment joining color(green)(""(5,1)) and color(purple)(""(5,10/3)) In fact the line segment between color(green)(""(5,1)) and any point on the line parallel to $\textcolor{red}{- 6 y + 3 x = 2}$ through color(purple)(""(5,10/3)) will be bisected by $\textcolor{red}{- 6 y + 3 x = 2}$

$\textcolor{red}{- 6 y + 3 x = 2}$ has a slope of $\textcolor{red}{\frac{1}{2}}$
so the line through color(purple)(""(5,10/3)) will also have a slope of $\textcolor{red}{\frac{1}{2}}$
and using the slope-point form, it will have an equation of
$\textcolor{w h i t e}{\text{XXX}} y - \textcolor{p u r p \le}{\frac{10}{3}} = \textcolor{red}{\frac{1}{2}} \left(x - \textcolor{p u r p \le}{5}\right)$
which can be simplified as
$\textcolor{w h i t e}{\text{XXX}} y = \frac{x}{2} + \frac{5}{6}$ 