# A line segment is bisected by a line with the equation  5 y -4 x = 1 . If one end of the line segment is at (3 ,8 ), where is the other end?

Oct 30, 2016

The other end is at the point $\left(7 , 3\right)$

#### Explanation:

It is implied that the bisector is perpendicular to bisected line, therefore we can use the slope of the bisector to find the slope of the bisected line.

Write the equation of the bisector line in slope-intercept form so that we may observe the slope:

$y = \frac{4}{5} x + 1$ [1]

The slope $m = \frac{4}{5}$

The slope, n, of the bisected line is the negative reciprocal, $n = - \frac{5}{4}$

Substitute the slope and the point, $\left(3 , 8\right)$, and the slope, $- \frac{5}{4}$ into the slope-intercept form of the equation of a line, $y = n x + b$ and then solve for b:

$8 = - \frac{5}{4} \left(3\right) + b$

$b = 8 + \frac{15}{4}$

$b = \frac{47}{4}$

The equation of the bisected line is:

$y = - \frac{5}{4} x + \frac{47}{4}$ [2]

Subtract equation [1] from equation [2]:

$y - y = - \frac{5}{4} x - \frac{4}{5} x + \frac{47}{4} - 1$

$0 = - \frac{5}{4} x - \frac{4}{5} x + \frac{47}{4} - 1$

$0 = - \frac{41}{20} x + \frac{43}{4}$

$\frac{41}{20} x = \frac{43}{4}$

$x = \frac{215}{43}$

The x coordinate of intersection is:

$x = 5$

The point of intersection is 2 units to the right of the starting point; the ending point must be an additional 2 units:

$x = 7$

Substitute 7 for x into equation [2]:

$y = - \frac{5}{4} \left(7\right) + \frac{47}{4}$

$y = - \frac{5}{4} \left(7\right) + \frac{47}{4}$

$y = 3$