# A line segment is bisected by a line with the equation  7 y + x = 7 . If one end of the line segment is at (1 ,3 ), where is the other end?

Jul 9, 2017

The other end is $= \left(0.4 , - 1.2\right)$

#### Explanation:

Let's rewrite the equation of the line.

$7 y + x = 7$

$7 y = - x + 7$

$y = - \frac{1}{7} x + 1$................................$\left(1\right)$

The slope is $m = - \frac{1}{7}$

The slope of the perpendicular line is $m ' = - \frac{1}{m} = - \frac{1}{- \frac{1}{7}} = 7$

The equation of the perpendicular line is

$y - 3 = 7 \left(x - 1\right)$

$y = 7 x - 7 + 3 = 7 x - 4$.......................$\left(2\right)$

The point of intersection is obtained by solving for $x$ and $y$ in equations $\left(1\right)$ and $\left(2\right)$

$7 x - 4 = - \frac{1}{7} x + 1$

$7 x + \frac{1}{7} x = 4 + 1 = 5$

$\frac{50}{7} x = 5$

$x = 5 \cdot \frac{7}{50} = \frac{7}{10} = 0.7$

$y = 7 \cdot 0.7 - 4 = 4.9 - 0.4 = 0.9$

The point of intersection is $= \left(0.7 , 0.9\right)$

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

Then,

$\left(\frac{a + 1}{2} , \frac{b + 3}{2}\right) = \left(0.7 , 0.9\right)$

$a + 1 = 2 \cdot 0.7 = 1.4$, $\implies$ , $a = 1.4 - 1 = 0.4$

$b + 3 = 2 \cdot 0.9 = 1.8$, $\implies$, $b = 1.8 - 3 = - 1.2$

The other end is $= \left(0.4 , - 1.2\right)$