# A line segment is bisected by a line with the equation  - 5 y + 3 x = 1 . If one end of the line segment is at (6 ,4 ), where is the other end?

Apr 29, 2016

$- 5 y + 3 x = 1$ is the bisector for a line segment between $\left(6 , 4\right)$ and any point on the line $\textcolor{b l u e}{- 5 y + 3 x = 4}$

#### Explanation: If $\left(x , y\right)$ is the end point of a line segment with $\left(6 , 4\right)$ as its other end; and
if $- 5 y + 3 x = 1$ bisects this line

Then for any point $\left(\overline{x} , \overline{y}\right)$ on $- 5 y + 3 x = 1$
the $\Delta x$ and $\Delta y$ between $\left(6 , 4\right)$ and $\left(\overline{x} , \overline{y}\right)$
will be the same as $\Delta x$ and $\Delta y$ between $\left(\overline{x} , \overline{y}\right)$ and the corresponding target end point.

Since $\Delta x = 6 - \overline{x}$ and $\Delta y = 4 - \overline{y}$
for any point $\left(\overline{x} , \overline{y}\right)$ on $- 5 y + 3 x = 1$
the corresponding target end point will be $\left(2 \overline{x} - 6 , 2 \overline{y} - 4\right)$

Specifically we can see that $\left(\overline{x} , \overline{y}\right) = \left(2 , 1\right)$ is a point on $- 5 y + 3 x = 1$
with a corresponding target point of $\left(- 2 , - 2\right)$

Note also that the target line is parallel to $- 5 y + 3 x = 1$
and therefore has the same slope (namely $\frac{3}{5}$).

Using the slope-point form for the target line, we get
$\textcolor{w h i t e}{\text{XXX}} \left(y + 2\right) = \frac{3}{5} \left(x + 2\right)$
or
$\textcolor{w h i t e}{\text{XXX}} 5 y + 10 = 3 x + 6$
or
$\textcolor{w h i t e}{\text{XXX}} - 5 y + 3 x = 4$