Can we find the point midway between an intersection and a line?

Can we find the point midway between an intersection and a line?

Question

$P$ is the point of intersection of the lines $y+x=1$ and $3y+2x=1$. Find the co-ordinates of the point midway between $P$ and the line $y=2x+8$.

Suggestion

Have you drawn a diagram yet? If you’re not sure where to start, sketch the three lines.

What is the gradient of a line perpendicular to $y=2x+8$? Can you find the equation of the one such line through $P$?

Solution

Let’s first find the coordinates of the point $P$ by solving the simultaneous equations $y+x=1$ and $3y+2x=1$.

We can use substitution: we know that at

$P$

,

$y=1-x$

, so

$$\begin{align*} &&3(1-x)+2x&=1,&&\quad \\ \implies&&3-3x+2x&=1,&& \\ \implies&&2&=x,&& \end{align*}$$

and so $P$ is the point $(2,1-2)=(2,-1)$.

Alternatively, we could have used elimination: multiplying

$y+x=1$

by

$2$

gives the simultaneous equations

$$\begin{align*} 2y+2x&=2\\ 3y+2x&=1. \end{align*}$$

Then either by subtracting these equations or by observation, we see that $y=-1$, so $x=2$, and $P$ is the point $(2,-1)$.

Now let’s draw a diagram showing $P$ and the line $y=2x+8$.

If we can find the line perpendicular to $y=2x+8$ through $P$, then we can find where these lines intersect, which we shall call $Q$. We can then compute the coordinates of the point midway between $Q$ and $P$, which we shall call $M$. This point $M$ is the point the question has asked us to find.

A line perpendicular to $y=2x+8$ must have gradient $-\frac12$. Therefore the perpendicular passing through $P(2,-1)$ has equation $y+1=-\tfrac{1}{2}(x-2)$ (using the general formula $y-y_1=m(x-x_1)$), which we can rearrange to get $y=-\frac12 x$.

So $Q$ is the point of intersection of $y=2x+8$ and $y=-\frac{1}{2}x$. Solving these simultaneously, we find that $-\tfrac{1}{2}x=2x+8,$ and so $\frac{5}{2}x=-8$, giving $x=-\frac{16}{5}$. Then we calculate $y=-\frac{1}{2}\times\left(-\frac{16}{5}\right)=\frac{8}{5},$ therefore $Q$ has coordinates $\bigl(-\frac{16}{5},\frac{8}{5}\bigr)$.

The coordinates of

$M$

are the average (mean) of the coordinates of

$P$

and

$Q$

. We therefore have

$$\begin{align*} M&=\left(\frac{2+\bigl(-\frac{16}{5}\bigr)}{2},\frac{-1+\frac{8}{5}}{2}\right)\\ &=\left(\frac{\bigl(-\frac{6}{5}\bigr)}{2},\frac{\bigl(\frac{3}{5}\bigr)}{2}\right)\\ &=\left(-\frac{3}{5},\frac{3}{10}\right). \end{align*}$$