# A line segment is bisected by line with the equation  3 y - 3 x = 1 . If one end of the line segment is at (2 ,5 ), where is the other end?

May 12, 2018

$\left(\frac{14}{3} , \frac{7}{3}\right)$

#### Explanation:

The perpendicular to

$- 3 x + 3 y = 1$

through $\left(2 , 5\right)$ is

$3 x + 3 y = 3 \left(2\right) + 3 \left(5\right) = 21$

Let's find where these meet. Subtracting equations,

$6 x = 20$

$x = \frac{10}{3}$

$6 y = 22$

$y = \frac{11}{3}$

So one endpoint is $A \left(2 , 5\right)$ and the bisection point is $B \left(\frac{10}{3} , \frac{11}{3}\right) .$ The other endpoint $C$ satisfies:

$C - B = B - A$

$C = 2 B - A = \left(2 \left(\frac{10}{3}\right) - 2 , 2 \left(\frac{11}{3}\right) - 5\right) = \left(\frac{14}{3} , \frac{7}{3}\right)$

Check:

The midpoint is

B = 1/2 (A+C) = ( (2+14/3)/2, (5+7/3)/2) = (10/3, 11/3) quad sqrt

Check the midpoint $B$ is on $3 y - 3 x = 1$

3( 11/3) -3 (10/3) = 1 quad sqrt

Check perpendicularity. The y intercept is $D \left(0 , \frac{1}{3}\right)$.

Let's check $\left(D - B\right) \cdot \left(C - A\right)$

$\left(0 - \frac{10}{3} , \frac{1}{3} - \frac{11}{3}\right) \cdot \left(\frac{14}{3} - 2 , \frac{7}{3} - 5\right)$

= (0 - 10/3) ( 14/3-2) + ( 7/3-5)( 1/3 - 11/3) = 0 quad sqrt

Zero dot product means perpendicularity.