# A line segment is bisected by a line with the equation  4 y - 3 x = 2 . If one end of the line segment is at ( 2 , 1 ), where is the other end?

Jan 24, 2017

$\left(1 \frac{1}{25} , 2 \frac{7}{25}\right)$

#### Explanation:

$4 y - 3 x = 2$, $4 y = 3 x + 2$
$y = \frac{3}{4} x + \frac{1}{2}$ $\to a$

gradient of the other line, $m = - \frac{1}{\frac{3}{4}} = - \frac{4}{3}$

The equation of the line thru $\left(2 , 1\right)$,

$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - 1 = - \frac{4}{3} \left(x - 2\right)$
$y = - \frac{4}{3} x + \frac{8}{3} + 1$
$y = - \frac{4}{3} x + \frac{11}{3}$

replacing y from $a$

$\frac{3}{4} x + \frac{1}{2} = - \frac{4}{3} x + \frac{11}{3}$

$\frac{3}{4} x + \frac{4}{3} x = \frac{11}{3} - \frac{1}{2}$

$\frac{9}{12} x + \frac{16}{12} x = \frac{22}{6} - \frac{3}{6}$

$\frac{25}{12} x = \frac{19}{6}$

$x = \frac{19}{6} \cdot \frac{12}{25} = \frac{38}{25}$

$y = \frac{3}{4} \left(\frac{38}{25}\right) + \frac{1}{2}$

$y = \frac{57}{50} + \frac{25}{50} = \frac{82}{50} = \frac{41}{25}$

$\left(\frac{38}{25} , \frac{41}{25}\right)$ is a midpoint of the line.

The other end of the line, $\left(x , y\right)$.

$\frac{x + 2}{2} = \frac{38}{25}$
$x = \frac{38}{25} \cdot 2 - 2$
$x = \frac{76}{25} - \frac{50}{25} = \frac{26}{25} = 1 \frac{1}{25}$

$\frac{y + 1}{2} = \frac{41}{25}$
$y = \frac{41}{25} \cdot 2 - 1$
$y = \frac{82}{25} - \frac{25}{25} = \frac{57}{25} = 2 \frac{7}{25}$

The other end $\left(1 \frac{1}{25} , 2 \frac{7}{25}\right)$