# A line segment is bisected by a line with the equation  - 6 y + 5 x = 4 . If one end of the line segment is at ( 2 , 5 ), where is the other end?

Dec 31, 2017

coordinates of the other end point $\textcolor{p u r p \le}{\frac{362}{61} , \frac{17}{61}}$

#### Explanation:

Assumption : Bisecting line is a perpendicular bisector

Standard form of equation $y = m x + c$
Slope of perpendicular bisector m is given by
$- 6 y + 5 x = 4$
$y = \left(\frac{5}{6}\right) x - \left(\frac{2}{3}\right)$
$m = \left(\frac{5}{6}\right)$

Slope of line segment is
$y - 5 = - \left(\frac{1}{m}\right) \left(x - 2\right)$
$y - 5 = \left(- \frac{6}{5}\right) \left(x - 2\right)$
$5 y - 25 = - 6 x + 12$

5y + 6x = 37 color (white)((aaaa) Eqn (1)
-6y + 5x = 4 color (white)((aaaa) Eqn (2)

Solving Eqns (1) & (2),

$x = \textcolor{g r e e n}{\frac{242}{61}}$

$y = \textcolor{g r e e n}{\frac{161}{61}}$

Mid point $\textcolor{g r e e n}{\frac{242}{61} , \frac{161}{61}}$

Let (x1,y1) the other end point.
$\frac{2 + x 1}{2} = \frac{242}{37}$
$x 1 = \left(\frac{484}{61}\right) - 2 = \textcolor{red}{\frac{362}{61}}$
$\frac{5 + y 1}{2} = \frac{161}{61}$
$y 1 = \left(\frac{322}{61}\right) - 5 \textcolor{red}{\frac{17}{61}}$

Coordinates of other end point $\textcolor{red}{\frac{362}{61} , \frac{17}{61}}$