# A line segment is bisected by line with the equation  6 y - 7 x = 3 . If one end of the line segment is at (7 ,2 ), where is the other end?

Oct 9, 2017

Other end point is $\left(\frac{103}{51} , \frac{130}{17}\right)$

#### Explanation:

Assumption : It’s a perpendicular bisector.
$6 y - 7 x = 3$ Eqn (1)
$y = \left(\frac{7}{6}\right) x + \cancel{\frac{3}{6}} \left(\frac{1}{2}\right)$
It’s in the form $y = m x + c$ where m is the slope.
:.m=7/6. Slope of perpendicular line is -(6/7)
Equation of line segment is
$y - 2 = - \left(\frac{6}{7}\right) \left(x - 7\right)$
$7 y - 14 = - 6 x + 42$
$7 y + 6 x = 56$ Eqn (2)

Solving Eqns (1) & (2)
$36 y - 42 x = 18$
$49 y + 42 x = 392$ Adding,
$85 y = 410$
$y = \frac{410}{85} = \frac{82}{17}$
x=(56-7y)/6=(56-(6*82)/17))/6=(952-492)/102=460/102=230/51

Mid point =(230/51,82/17) One end point =(7,2) Other end point be (x1,y1)
$\frac{x 1 + 7}{2} = \frac{230}{51}$
$x 1 = \left(\frac{460}{51}\right) - 7 = \frac{460 - 357}{51} = \frac{103}{51}$
$\frac{y 1 + 2}{2} = \frac{82}{17}$
$y 1 = \left(\frac{164}{17}\right) - 2 = \frac{164 - 34}{17} = \frac{130}{17}$
Other end point is $\left(\frac{103}{51} , \frac{130}{17}\right)$