A line segment is bisected by a line with the equation  4 y + 9 x = 8 . If one end of the line segment is at (5 ,2 ), where is the other end?

Oct 19, 2016

The other end is at $\left(\frac{1295}{97} , \frac{554}{97}\right)$

Explanation:

Write the given line in slope-intercept form:

$y = - \frac{9}{4} x + 2$

Because bisector is perpendicular, the slope of the line segment will be the negative reciprocal of its bisector, $\frac{4}{9}$.

This reference gives us an equation for the distance from the point to the line. The distance from the point to the line is:

$d = | \frac{4 \left(2\right) + 9 \left(5\right) - 8}{\sqrt{{4}^{2} + {9}^{2}}} | = 45 \frac{\sqrt{97}}{97}$

The length of the line segment is twice this distance, $90 \frac{\sqrt{97}}{97}$.

From point $\left(5 , 2\right)$, we move to the right a distance, (x), and up a distance y , we know that y is $\frac{4}{9} x$, and we know that length of the hypotenuse formed by this right triangle is $90 \frac{\sqrt{97}}{97}$

${\left(90 \frac{\sqrt{97}}{97}\right)}^{2} = {x}^{2} + {\left(\frac{4}{9} x\right)}^{2}$

${90}^{2} / 97 = \frac{81}{81} {x}^{+} \frac{16}{81} {x}^{2}$

${x}^{2} = 81 \frac{{90}^{2}}{97} ^ 2$

$x = \frac{810}{97}$

$y = \frac{360}{97}$

The other end of the line is at point:

$\left(5 + \frac{810}{97} , 2 + \frac{360}{97}\right) = \left(\frac{1295}{97} , \frac{554}{97}\right)$