# A line segment is bisected by a line with the equation  -6 y - x = 3 . If one end of the line segment is at ( 8 , 3 ), where is the other end?

Oct 4, 2017

The other end point is $\left(\frac{238}{37} , - \frac{237}{37}\right)$

#### Explanation:

It is assumed that the line bisecting the line segment is assumed to be a perpendicular bisector.
$- 6 y - x = 3$ Eqn (1)
$- 6 y = x + 3$
$y = - \left(\frac{x}{6}\right) - \left(\frac{1}{2}\right)$
Slope of the equation $m 1 = - \left(\frac{1}{6}\right)$
Slope the line segment $m 2 = - \left(\frac{1}{m 1}\right) = - \left(\frac{1}{- \frac{1}{6}}\right) = 6$
Equation of the line segment is
$\left(y - 3\right) = 6 \cdot \left(x - 8\right)$
$y - 6 x = - 45$ Eqn (2)

Solving equations (1) & (2), we get the midpoint which is also the midpoint of the line segment.
$- x - 6 y = 3$Eqn (1)
$- 36 x + 6 y = - 270$ Eqn (2) * 6; Adding,
$- 37 x = - 267$
$x = \frac{267}{37}$
Substituting value of x in Eqn (1),
$- 6 y = 3 + \left(\frac{267}{37}\right)$
y=-(111+267)/(37*6)=-(378/(222)=-(63/3)
Coordinate of midpoint$\left(\frac{267}{37}\right) , - \left(\frac{63}{37}\right)$
Let (x2,y2) be the other end point coordinates of the line segment.
$\frac{x 2 + 8}{2} = \left(\frac{267}{37}\right)$
$x 2 = \left(\frac{534}{37}\right) - 8 = \frac{534 - 296}{37} = \frac{238}{37}$
$\frac{y 2 + 3}{2} = - \left(\frac{63}{37}\right)$
y2+3=-(126/37) y2=-(126+111)/37=-(237/37)#