# A line segment is bisected by a line with the equation  9 y - 2 x = 5 . If one end of the line segment is at ( 7 , 3 ), where is the other end?

Jun 22, 2018

The general line containing all the endpoints of line segments with other endpoint $\left(7 , 3\right)$ bisected by $9 y - 2 x = 5$ is

$9 y - 2 x = - 3$

The endpoint of the perpendicular bisector is $\left(\frac{627}{85} , \frac{111}{85}\right) .$

#### Explanation:

If the line is the perpendicular bisector of the segment, the other endpoint is uniquely determined.

If the line is just a bisector, as asked, each point on the line is the bisector of some segment whose endpoint is $\left(7 , 3\right)$ and whose other endpoint sweeps out a curve. Let's see if we can figure out its equation.

If we're told $\left(p , q\right)$ is the midpoint of a segment with endpoint $\left(a , b\right) = \left(7 , 3\right)$ the other endpoint $\left(x , y\right)$ satisfies:

$\left(x , y\right) - \left(p , q\right) = \left(p , q\right) - \left(a , b\right)$

$\left(x , y\right) = \left(2 p - a , 2 q - b\right)$

$\left(x , y\right) = \left(2 p - 7 , 2 q - 3\right)$

We have $\left(p , q\right)$ on the line so $2 p = 9 q - 5$.

$\left(x , y\right) = \left(9 q - 12 , 2 q - 3\right) = \left(- 12 , - 3\right) + q \left(9 , 2\right)$

The other endpoint $\left(x , y\right)$ makes a line through $\left(- 12 , - 3\right)$ with direction vector $\left(9 , 2\right)$ meaning slope $\frac{2}{9.}$ Nonparametrically, that's the line

$y + 3 = \frac{2}{9} \left(x + 12\right)$

$9 y + 27 = 2 x + 24$

$9 y - 2 x = - 3$

That's parallel to the original line, as far from it as $\left(7 , 3\right)$ is.

Now let's ask which $\left(x , y\right)$ makes a perpendicular bisector to the line?

The direction of the segment is $\left(x - a , y - b\right) = \left(x - 7 , y - 3\right) .$ The direction of the line $- 2 x + 9 y = 5$ is $\left(9 , 2\right)$ meaning for every $9$ units $x$ increases we increase $y$ by $2$ units.

We have perpendicularity when the dot product is zero:

$\left(x - 7 , y - 3\right) \cdot \left(9 , 2\right) = 0$

$9 \left(9 q - 12 - 7\right) + 2 \left(2 q - 3 - 3\right) = 0$

$85 q - 183 = 0$

$q = \frac{183}{85}$

$\left(x , y\right) = \left(9 q - 12 , 2 q - 3\right) = \left(\frac{627}{85} , \frac{111}{85}\right)$

Check:

 ( (627/85,111/85)-(7,3) ) cdot (9,2) = 0 quad sqrt