# A line segment is bisected by a line with the equation  - y + 3 x = 1 . If one end of the line segment is at (6 ,3 ), where is the other end?

May 24, 2018

color(blue)((-12/5,29/5)

#### Explanation:

We know that the line $- y + 3 x = 1$ and the line containing the point $\left(6.3\right)$ are perpendicular. If two lines are perpendicular then the product of their gradients is $- 1$

$- y + 3 x = 1 \implies y = 3 x - 1 \setminus \setminus \setminus \left[1\right]$

This has a gradient of 3. The line containing $\left(6 , 3\right)$ therefore has a gradient:

$m \cdot 3 = - 1 \implies m = - \frac{1}{3}$

Finding the equation of this line using point slope method:

$y - 3 = - \frac{1}{3} \left(x - 6\right) \implies y = - \frac{1}{3} x + 5 \setminus \setminus \setminus \setminus \left[2\right]$

Solving $\left[1\right] \mathmr{and} \left[2\right]$ simultaneously:

$- \frac{1}{3} x + 5 - 3 x + 1 = 0 \implies x = \frac{9}{5}$

Substitute in $\left[1\right]$

$y = 3 \left(\frac{9}{5}\right) - 1 = \frac{22}{5}$

$\left(\frac{9}{5} , \frac{22}{5}\right)$ are the coordinates of the midpoint.

Coordinates of the midpoint are found using:

$\left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$

Let the unknown endpoint be $\left({x}_{2} , {y}_{2}\right)$

Then:

$\left(\frac{6 + {x}_{2}}{2} , \frac{3 + {y}_{2}}{2}\right) \to \left(\frac{9}{5} , \frac{22}{5}\right)$

$\frac{6 + {x}_{2}}{2} = \frac{9}{5} \implies {x}_{2} = - \frac{12}{5}$

$\frac{3 + {y}_{2}}{2} = \frac{22}{5} \implies {y}_{2} = \frac{29}{5}$

Coordinates of the other endpoint are:

$\left(- \frac{12}{5} , \frac{29}{5}\right)$