Challenge Level

We received correct solutions from Ash and
Lucy, two students at Tiffin Girls' School. Well done to you
both.

Ash sent us this correct solution:

The gradient of the first line must be the negative reciprocal
of the gradient of the other line:

if you multiply the two gradients you always get -1.

For example:

Take a random gradient, say $\frac{4}{7}$

The negative reciprocal gradient will be $\frac{-7}{4}$

Make up two equations with these gradients, say

$y =\frac{4x}{7}$ and $y =\frac{-7x}{4}$

Draw them on a grid

You get perpendicular lines.

Lucy recorded how she worked through this
problem:

$y = x$ is perpendicular to $y = -x + 2$

(the y-intercept doesn't affect the
gradient of the line)

$y = 2x$ is perpendicular to $y = -x/2$

$y = -3x$ is perpendicular to $y = x/3 $

(to make a line perpendicular you
need to invert the gradient, or take the reciprocal, and change the
sign)

$y = -2x$ is perpendicular to $y = x/2 $

I can see a pattern here: when the two gradients of perpendicular lines are multiplied together they give -1, and the y-intercept does not affect if the line is perpendicular or not.

I will now try to work out what the perpendicular line of some other lines will be using this formula:

$y = 7x - 3$:

Using my formula I predict that a line which is perpendicular to this line will be $y = -x/7 - 3$

When I tested the lines out, I found that the formula had worked.

$y = x/3 + 4$:

Using my formula I predict that a line which is perpendicular to this line will be $y = -3x + 4$

Having drawn out the lines I found that the formula worked and the lines were perpendicular.

$y = -7x/3 + 2$:

I predict that a line which is perpendicular to this line will be $y = 3x/7$

From drawing out these lines I can see that they are perpendicular.