# A line segment is bisected by a line with the equation  -6 y + 9 x = 2 . If one end of the line segment is at ( 5 , 1 ), where is the other end?

##### 1 Answer
Oct 22, 2016

The point is $\left(- \frac{9}{13} , \frac{187}{39}\right)$

#### Explanation:

Write the given equation in slope-intercept form:

$y = \frac{3}{2} x - \frac{1}{3}$

The slope is $\frac{3}{2}$ any line perpendicular will have $- \frac{2}{3}$ slope.

Use the point-slope form of the equation of a line, to find the equation of the bisected line:

$y - 1 = - \frac{2}{3} \left(x - 5\right)$

$y = - \frac{2}{3} x + \frac{13}{3}$

Subtract the second line from the first:

$0 = \left(\frac{3}{2} + \frac{2}{3}\right) x - \frac{14}{3}$

$\frac{13}{6} x = \frac{14}{3}$

$x = \frac{28}{13}$

The change in x from $5$ to $\frac{28}{13}$ is:

$\frac{28}{13} - \frac{13}{13} \left(5\right) = - \frac{37}{13}$

Add twice that number to 5 to get the x coordinate of the other end of the line segment:

13/13(5) - 74/13 = -9/13

Substitute into the equation of the line for the y coordinate:

$y = - \frac{2}{3} \left(- \frac{9}{13}\right) + \frac{13}{3}$

$y = \frac{18}{39} + \frac{169}{39}$

$y = \frac{187}{39}$