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

Nov 23, 2017

$Q = \left(\frac{17}{5} , - \frac{1}{5}\right)$

#### Explanation:

First off, even if it is crude, draw a sketch.

If the bisectors are perpendicular, then the products of their gradients will be $- 1$

Let $L$ be $2 y + x = 7$
$y = - \frac{1}{2} x + \frac{7}{2}$
I will be using $m$ to denote the gradient (its easier than writing grad for formatting)
$\therefore {m}_{L} = - \frac{1}{2}$
Let $B =$bisector
${m}_{B} \cdot {m}_{L} = - 1$
$\therefore {m}_{B} = 2$

Now, we are searching for the equation of a line given a gradient and a point. This will allow us to easily find points on the bisector. We will use:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$ where $\left({x}_{1} , {y}_{1}\right)$ is a co-ordinate and $m$ is the gradient.

$\therefore$ eqn of B;

$y - 3 = 2 \left(x - 5\right)$
$y - 3 = 2 x - 10$
$y = 2 x - 7$

Going back to our sketch, there is a point that these two lines cross. We can find this point by solving simultaneously for B and L

Subst B into L

$2 \left(2 x - 7\right) + x = 7$
$5 x - 14 = 7$
$5 x = 21$
$x = \frac{21}{5}$

$y = 2 x - 7$
$y = 2 \cdot \frac{21}{5} - 7$
$y = \frac{7}{5}$

Let $M = \left(\frac{21}{5} , \frac{7}{5}\right)$

We can now use column vectors to get us to Q. Since the part of the segment either side of the line bisecting it is equal, we know that the vector $\vec{P M} = \vec{M Q}$

$\vec{P M} = \left(\begin{matrix}\frac{21}{5} - 5 \\ \frac{7}{5} - 3\end{matrix}\right)$
$= \left(\begin{matrix}- \frac{4}{5} \\ - \frac{8}{5}\end{matrix}\right)$
$\therefore \vec{M Q} = \left(\begin{matrix}- \frac{4}{5} \\ - \frac{8}{5}\end{matrix}\right)$
$Q = \left(\frac{21}{5} - \frac{4}{5} , \frac{7}{5} - \frac{8}{5}\right)$
$Q = \left(\frac{17}{5} , - \frac{1}{5}\right)$

There is quite probably a simpler way to do this, but I don't know of any. Please tell me if I need to clarify anything