# A line segment is bisected by a line with the equation  4 y - 2 x = 3 . If one end of the line segment is at ( 5 , 6 ), where is the other end?

Jun 16, 2018

The other end of the line segment is at $\left(7.2 , 1.6\right)$

#### Explanation:

First of all, the bisector of a line segment is perpendicular to the segment.

Perpendicular slope is the negative reciprocal of the original slope, so first step is to find the slope of the given line.

Step 1: Finding the equation of the line segment

Convert to slope-intercept form $y = m x + b$

$4 y - 2 x = 3$
$4 y = 2 x + 3$
$y = \frac{2}{4} x + \frac{3}{4}$
$\textcolor{red}{m = \frac{2}{4}}$

Taking the negative reciprocal we have the slope of our line segment $\textcolor{red}{m = - \frac{4}{2}}$

NOTE: Because the bisector intersects the line segment at the midpoint, we can use the given point to find our $b$ value.

$6 = - \frac{4}{2} \left(5\right) + b$
$6 = - \frac{20}{2} + b$
$6 + 10 = b$
$\textcolor{red}{b = 16}$

This gives us the equation of the line segment on infinite domain $y = - \frac{4}{2} x + 16$

Step 2: Equate the two lines to find the intersection point at x

$\frac{2}{4} x + \frac{3}{4} = - \frac{4}{2} x + 16$

Group x values together and factor

$x \left(\frac{2}{4} + \frac{4}{2}\right) = - \frac{3}{4} + 16$
$x \left(\frac{2}{4} + \frac{8}{4}\right) = - \frac{3}{4} + \frac{64}{4}$

$\frac{10 x}{4} = \frac{61}{4}$

$10 x = 61$

$\textcolor{red}{x = 6.1}$

Sub in 6.1 to either equation to find y.

$y = - \frac{4}{2} \left(6.1\right) + 16$
$y = - \frac{24.4}{2} + 16$
$y = - 12.2 + 16$
$y = 3.8$

So our intersection point is at $\left(6.1 , 3.8\right)$

Step 3: Find the endpoint using the midpoint formula

Because the intersection point $\left(6.1 , 3.8\right)$ is also a midpoint, we can find the final point using the midpoint formula.

$\left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$

Substitute the original points at ${x}_{1} , {y}_{1}$

$\frac{5 + {x}_{2}}{2} = 6.1$
${x}_{2} = 6.1 \left(2\right) - 5$
${x}_{2} = 7.2$

$\frac{6 + {y}_{2}}{2} = 3.8$
${y}_{2} = 3.8 \left(2\right) - 6$
${y}_{2} = 1.6$

$\left(x , y\right) = \left(7.2 , 1.6\right)$ 