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

Jul 15, 2016

The other end could be any point on the line $y = 2 x - 1$

#### Explanation:

For convenience, I will rearrange the given equation
$\textcolor{w h i t e}{\text{XXX}} - 3 y + 6 x = 6$
as
$\textcolor{w h i t e}{\text{XXX}} y = 2 x - 2$

Consider the vertical line through $\left(3 , 3\right)$.
Since $x$ is constant for all points on a vertical line,
this vertical line will intersect $y = 2 x - 2$ at $\left(3 , 4\right)$

The distance between $\left(3 , 3\right)$ and $\left(3 , 4\right)$ is $1$ unit.

The point $\left(3 , 5\right)$ is also on this vertical line at a distance of $2$ units from $\left(3 , 3\right)$.

Therefore $y = 2 x - 2$ (originally given as $- 3 y + 6 x = 6$) bisects the line segment joining $\left(3 , 3\right)$ and $\left(3 , 5\right)$.

Therefore one possible endpoint would be at $\left(3 , 5\right)$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Perhaps less obviously, any point on a line through $\left(3 , 5\right)$ and parallel to the given line $y = 2 x - 2$ (or in its original but less convenient form $- 3 y + 6 x = 6$)
will also be a line segment endpoint bisected by the given equation. From the image above we can see that for an arbitrary point $E$ on a line parallel to the given bisector line
triangle ABD ~ triangle ACE
and since $\left\mid A B \right\mid = \frac{1}{2} \left\mid A C \right\mid$
$\rightarrow \left\mid A D \right\mid = \frac{1}{2} \left\mid A E \right\mid$
(i.e. the given line is a bisector for $A E$)

Since $y = 2 x - 2$ has a slope of $2$
the line parallel to it and through $\left(3 , 5\right)$ will also have a slope of $2$
and using the slope-point form:
$\textcolor{w h i t e}{\text{XXX}} y - 5 = 2 \left(x - 3\right)$
Which can be simplified as
$\textcolor{w h i t e}{\text{XXX}} y = 2 x - 1$