# A line segment is bisected by a line with the equation  - 5 y + 2 x = 1 . If one end of the line segment is at (6 ,4 ), where is the other end?

##### 1 Answer
Jul 14, 2017

The coordinates of the other end is $= \left(\frac{210}{29} , \frac{28}{29}\right)$

#### Explanation:

The equation of the line is

$- 5 y + 2 x = 1$

$5 y = 2 x - 1$

$y = \frac{2}{5} x - \frac{1}{5}$.......................$\left(1\right)$

The slope of the line is $m = \frac{2}{5}$

The slope of the segment is $m '$

$m m ' = - 1$

$m ' = - \frac{5}{2}$

The equation of the segment is

$y - 4 = - \frac{5}{2} \left(x - 6\right)$

$y = - \frac{5}{2} x + 15 + 4$

$y = - \frac{5}{2} x + 19$.......................$\left(2\right)$

Solving for $x$ and $y$ in the equations $\left(1\right)$ and $\left(2\right)$ gives the point of intersection of the line and the segment

$- \frac{5}{2} x + 19 = \frac{2}{5} x - \frac{1}{5}$

$\frac{2}{5} x + \frac{5}{2} x = 19 + \frac{1}{5}$

$\frac{29}{10} x = \frac{96}{5}$

$x = \frac{96}{5} \cdot \frac{10}{29} = \frac{192}{29}$

$y = \frac{2}{5} \cdot \frac{192}{29} - \frac{1}{5} = \frac{71}{29}$

Let the other end of the segment be $= \left(a , b\right)$

Therefore,

$\left(\frac{192}{29} , \frac{71}{29}\right) = \left(\frac{a + 6}{2} , \frac{b + 4}{2}\right)$

$\frac{a + 6}{2} = \frac{192}{29}$

$a = \frac{384}{29} - 6 = \frac{210}{29}$

$\frac{b + 4}{2} = \frac{71}{29}$

$b = \frac{144}{29} - 4 = \frac{28}{29}$