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

May 3, 2018

$\left(- \frac{91}{29} , \frac{213}{29}\right)$

#### Explanation:

Let's do a parametric solution, which I think is slightly less work.
Let's write the given line

$- 7 x + 3 y = 2 \quad \quad \quad \quad \quad \quad \quad \quad y = \frac{7}{3} x + \frac{2}{3}$

I write it this way with $x$ first so I don't accidentally substitute in a $y$ value for an $x$ value. The line has a slope of $\frac{7}{3}$ so a direction vector of $\left(3 , 7\right)$ (for every increase in $x$ by $3$ we see $y$ increase by $7$). This means the direction vector of the perpendicular is $\left(7 , - 3\right) .$

The perpendicular through $\left(7 , 3\right)$ is thus

$\left(x , y\right) = \left(7 , 3\right) + t \left(7 , - 3\right) = \left(7 + 7 t , 3 - 3 t\right)$.

This meets the original line when

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

$- 58 t = 42$

$t = - \frac{42}{58} = - \frac{21}{29}$

When $t = 0$ we're at $\left(7 , 3\right) ,$ one end of the segment, and when $t = - \frac{21}{29}$ we're at the bisection point. So we double and get $t = - \frac{42}{29}$ gives the other end of the segment:

$\left(x , y\right) = \left(7 , 3\right) + \left(- \frac{42}{29}\right) \left(7 , - 3\right) = \left(- \frac{91}{29} , \frac{213}{29}\right)$

Check:

We check bisector then we check perpendicular.

The midpoint of the segment is

$\left(\frac{7 + - \frac{91}{29}}{2} , \frac{3 + \frac{213}{29}}{2}\right) = \left(\frac{56}{29} , \frac{150}{29}\right)$

We check that's on $- 7 x + 3 y = 2$

 - 7 (56/29) + 3 ( 150/29) = 2 quad sqrt

Let's check it's a zero dot product of the difference of the segment endpoints with the direction vector $\left(3 , 7\right)$:

 3 (-91/29 - 7) + 7( 213/29 - 3) = 0 quad sqrt