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
$y = -x$
$y = x$ 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.