# A line segment is bisected by a line with the equation  3 y + 5 x = 2 . If one end of the line segment is at ( 5 , 8 ), where is the other end?

Nov 22, 2016

The other end is at $\left(- \frac{150}{17} , - \frac{5}{17}\right)$

#### Explanation:

This problem is in the perpendicular bisectors so it is safe to assume that the bisected line is perpendicular; that means that its equation is of the form:

$5 y - 3 x = c$

To find the value of c, substitute in the point $\left(5 , 8\right)$

$5 \left(8\right) - 3 \left(5\right) = c$

$c = 25$

The equation of the bisected line is:

$5 y - 3 x = 25 \text{ [1]}$

The given equation is:

$3 y + 5 x = 2 \text{ [2]}$

Multiply equation [1] by -3 and equation [2] by 5

$- 15 y + 9 x = - 75 \text{ [3]}$
$15 y + 25 x = 10 \text{ [4]}$

$34 x = - 65$

$x = - \frac{65}{34}$

Let $\Delta x$ = the change in x from the given point to the intersection point:

$\Delta x = - \frac{65}{34} - 5 = - \frac{235}{34}$

The x coordinate of the other end of the line segment is twice that change added to the starting x coordinate:

${x}_{e n d} = 2 \Delta x + 5$

${x}_{e n d} = 2 \left(- \frac{235}{34}\right) + 5$

${x}_{e n d} = - \frac{150}{17}$

To find the y coordinate, ${y}_{e n d}$, substitute $- \frac{150}{17}$ for x

$5 y - 3 \left(- \frac{150}{17}\right) = 25$

$5 y = 3 \left(- \frac{150}{17}\right) + 25$

${y}_{e n d} = 3 \left(- \frac{30}{17}\right) + 5$

${y}_{e n d} = - \frac{5}{17}$

The other end is at $\left(- \frac{150}{17} , - \frac{5}{17}\right)$