# A line segment is bisected by a line with the equation  4 y + 3 x = 4 . If one end of the line segment is at ( 8 , 9 ), where is the other end?

Oct 22, 2016

The other end is the point $\left(- \frac{168}{25} , - \frac{223}{25}\right)$

#### Explanation:

Write the equation of the bisector is slope-intercept form:

$y = - \frac{3}{4} x + 1 \textcolor{w h i t e}{_} \left[1\right]$

The slope of the bisected line is the negative reciprocal, $\frac{4}{3}$. Use the point-slope form of the equation of a line to force the line through point $\left(8 , 9\right)$

$y - 9 = \frac{4}{3} \left(x - 8\right)$

$y - 9 = \frac{4}{3} x - \frac{32}{3}$

$y = \frac{4}{3} x - \frac{32}{3} + 9$

$y = \frac{4}{3} x - \frac{5}{3} \textcolor{w h i t e}{_} \left[2\right]$

Subtract equation [1] from equation [2]:

$0 = \left(\frac{4}{3} + \frac{3}{4}\right) x - \frac{8}{3}$

Solve for x:

$\frac{8}{3} = \frac{25}{12} x$

This is the x coordinate of the point of intersection:

$x = \frac{32}{25}$

The change in x from the given point to the point of intersection:

$\Delta x = \frac{32}{25} - 8 = - \frac{168}{25}$

The x coordinate of the other end of the line twice the above added to 8:

$x = 8 + 2 \left(- \frac{168}{25}\right)$

$x = - \frac{136}{25}$

Substitute $x = - \frac{136}{25}$ into equation [2]:

$y = \frac{4}{3} \left(- \frac{136}{25}\right) - \frac{5}{3}$

$y = - \frac{544}{75} - \frac{125}{75}$

$y = - \frac{669}{75} = - \frac{223}{25}$

The other end is the point $\left(- \frac{168}{25} , - \frac{223}{25}\right)$