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

Jan 26, 2017

$\left(4 \frac{1}{2} , - 2 \frac{1}{2}\right)$

#### Explanation:

$6 y - 2 x = 1$
$6 y = 2 x + 1$

$y = \frac{1}{3} x + \frac{1}{6}$
The slope for this equation is $\frac{1}{3}$, therefor the slope for line segment,$m = - 3$ where $m \cdot {m}_{1} = - 1$

The equation of line segment is $\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$ where ${x}_{1} = 2 , {y}_{1} = 5$

$\left(y - 5\right) = - 3 \left(x - 2\right)$
$y = - 3 x + 6 + 5 = - 3 x + 11$ $\to a$

so, the intercept between 2 lines is
$\frac{1}{3} x + \frac{1}{6} = - 3 x + 11$

$\frac{2}{6} x + \frac{1}{6} = - 3 x + 11$

$2 x + 1 = 6 \left(- 3 x + 11\right)$
$2 x = - 18 x + 66 - 1$
$20 x = 65$
$x = \frac{65}{20} = \frac{13}{4} = 3 \frac{1}{4}$

therefore,
$y = - 3 \left(\frac{13}{4}\right) + 11$
$y = - \frac{39}{4} + \frac{44}{4}$
$y = \frac{5}{4} = 1 \frac{1}{4}$

The line which intercept with both lines is a midpoint of the line segment. Therefore the other end line $\left(x , y\right)$

$\frac{x + 2}{2} = \frac{13}{4}$, $\frac{y + 5}{2} = \frac{5}{4}$

$x + 2 = \frac{13}{2}$, $y + 5 = \frac{5}{2}$

$x = \frac{13}{2} - 2$, $y = \frac{5}{2} - 5$
$x = \frac{9}{2} = 4 \frac{1}{2}$, $y = - \frac{5}{2} = - 2 \frac{1}{2}$