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

Feb 25, 2018

The other end is at $\left(\frac{33}{17} , - \frac{106}{17}\right)$

#### Explanation:

Given the equation of the perpendicular bisector of the form:

$A x + B y = {C}_{1}$

The equation of the bisected line is of the form:

$B x - A y = {C}_{2}$

Writing the equation of the given perpendicular bisector in the above form:

$x + 4 y = 3$

Now, we can with the general form for the bisected line:

$4 x - y = {C}_{2}$

To find the value of ${C}_{2}$, we substitute the point $\left(5 , 6\right)$:

$4 \left(5\right) - 6 = {C}_{2}$

${C}_{2} = 14$

The equation of the bisected line is:

$4 x - y = 14$

Find the point of intersection of the lines:

$x + 4 y = 3$
$4 x - y = 14$

Multiply the second equation by 4 and add to the first equation:

$17 x = 59$

Solve for x:

$x = \frac{59}{17}$

Use the given line to solve for y:

$\frac{59}{17} + 4 y = 3$

$y = - \frac{2}{17}$

We can use the midpoint equations to find the other end:

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

y_"mid" = (y_"start"+y_"end")/2

Substitute $\left(5 , 6\right)$ for the starts and $\left(\frac{59}{17} , - \frac{2}{17}\right)$ for mids:

$\frac{59}{17} = \frac{5 + {x}_{\text{end}}}{2}$

$- \frac{2}{17} = \frac{6 + {y}_{\text{end}}}{2}$

Solve for $\left({x}_{\text{end", y_"end}}\right)$

${x}_{\text{end}} = \frac{33}{17}$

${y}_{\text{end}} = - \frac{106}{17}$

The other end is at $\left(\frac{33}{17} , - \frac{106}{17}\right)$