# A line segment is bisected by a line with the equation  - 3 y + 5 x = 8 . If one end of the line segment is at ( 7 , 9 ), where is the other end?

Dec 31, 2017

Coordinates of the other end point $\textcolor{b l u e}{- \frac{61}{17} , - \frac{147}{17}}$

#### Explanation:

Assumption : Bisecting line is a perpendicular bisector

Standard form of equation $y = m x + c$
Slope of perpendicular bisector m is given by
$- 3 y + 5 x = 8$
$y = \left(\frac{5}{3}\right) x - \left(\frac{8}{3}\right)$
$m = \left(\frac{5}{3}\right)$

Slope of line segment is
$y - 9 = - \left(\frac{1}{m}\right) \left(x - 7\right)$
$y - 9 = \left(- \frac{3}{5}\right) \left(x - 7\right)$
$5 y - 45 = - 3 x + 21$

-3y + 5x = 8 color (white)((aaaa) Eqn (1)
5y + 3x = 66 color (white)((aaaa) Eqn (2)

Solving Eqns (1) & (2),

$x = \frac{\textcolor{p u r p \le}{29}}{17}$

$y = \frac{\textcolor{p u r p \le}{3}}{17}$
Mid point $\textcolor{p u r p \le}{\frac{29}{17} , \frac{3}{17}}$

Let (x1,y1) the other end point.
$\frac{7 + x 1}{2} = \frac{29}{17}$
$x 1 = \left(\frac{58}{17}\right) - 7 = \textcolor{red}{- \frac{61}{17}}$
$\frac{9 + y 1}{2} = \frac{3}{17}$
$y 1 = \left(\frac{6}{17}\right) - 9 = \textcolor{red}{- \frac{147}{17}}$

Coordinates of other end point $\textcolor{red}{- \frac{61}{17} , - \frac{147}{17}}$