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

Oct 18, 2016

The other end will be any point on the line
$\textcolor{w h i t e}{\text{XXX}} - 3 y + 6 x = 1$

Explanation:

To find one point that satisfy the required conditions, we could take the horizontal line through $\left(3 , 3\right)$ and note that it intersects the given bisector equation at $\left(\frac{7}{3} , 3\right)$

$\left(\frac{7}{3} , 3\right)$ is $\frac{2}{3}$ to the left of $\left(3 , 3\right)$
The point horizontally $\frac{2}{3}$ further to the left of $\left(\frac{7}{3} , 3\right)$
is $\left(\frac{5}{3} , 3\right)$

The bisector line, $- 3 y + 6 x = 5$ bisects the line segment between $\left(3 , 3\right)$ and $\left(\frac{5}{3} , 3\right)$

Furthermore any point, $P$ on the line through $\left(\frac{5}{3} , 3\right)$ and parallel to the bisector line $- 3 y + 6 x = 5$
will provide an endpoint that meets the requirement.

$- 3 y + 6 x = 5$ has a slope of $3$
So the required line will have a slope of $3$ and pass through $\left(\frac{5}{3} , 3\right)$
Using the slope-point form and then manipulating the derived equation into a similar form to that of the bisector line,
we get $- 3 y + 6 x = 1$