# A line segment is bisected by a line with the equation  4 y - 2 x = 5 . If one end of the line segment is at ( 7 , 3 ), where is the other end?

May 13, 2017

The other end is $\left(\frac{28}{5} , \frac{29}{5}\right)$

#### Explanation:

Given: $4 y - 2 x = 5 \text{ [1]}$ is the equation of the bisector

Find the slope of the bisector:

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

$y = \frac{1}{2} x + \frac{5}{4}$

The slope is ${m}_{1} = \frac{1}{2}$

The of the bisected line is:

${m}_{2} = - \frac{1}{m} _ 1$

${m}_{2} = - \frac{1}{\frac{1}{2}}$

${m}_{2} = - 2$

Use the point-slope form of the equation of a line to find the equation of the bisected line:

$y = m \left(x - {x}_{0}\right) + {y}_{0}$

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

$y = - 2 x + 14 + 3$

$y = - 2 x + 17$

$y + 2 x = 17 \text{ [2]}$

Add equation [2] to equation [1]:

$4 y + y - 2 x + 2 x = 5 + 17$

$5 y = 22$

$y = \frac{22}{5}$

Use equation [2] to find the corresponding value of x:

$\frac{22}{5} + 2 x = 17$

$2 x = \frac{63}{5}$

$x = \frac{63}{10}$

Use the midpoint formulas:

x_"midpoint" = (x_"end"+x_"start")/2

y_"midpoint" = (y_"end"+x_"start")/2

Substitute $\frac{63}{10}$ for ${x}_{\text{midpoint}}$ and 7 for ${x}_{\text{start}}$:

$\frac{63}{10} = \frac{{x}_{\text{end}} + 7}{2}$

Substitute $\frac{22}{5}$ for ${y}_{\text{midpoint}}$ and 3 for ${y}_{\text{start}}$:

$\frac{22}{5} = \frac{{y}_{\text{end}} + 3}{2}$

Solve for ${x}_{\text{end" and y_"end}}$

$\frac{63}{10} = \frac{{x}_{\text{end}} + 7}{2}$
$\frac{22}{5} = \frac{{y}_{\text{end}} + 3}{2}$

${x}_{\text{end}} = \frac{63}{5} - 7$
${y}_{\text{end}} = \frac{44}{5} - 3$

${x}_{\text{end}} = \frac{28}{5}$
${y}_{\text{end}} = \frac{29}{5}$

The other end is $\left(\frac{28}{5} , \frac{29}{5}\right)$