13.12 Horizontal and vertical lines

We know that if there is a common difference between the $y$y-values as the $x$x-value changes by a constant amount, then there is a linear relationship. But what if there is no change in the $y$y-values at all? Or if the $y$y-values change but the $x$x-value remains the same?

Exploration

Consider the following table of values

We can see that as the $x$x-value increases by $1$1, the $y$y-value does not change at all. We can think of this as increasing, or decreasing for that matter, by $0$0 each step.

We know that in a linear equation of the form $y=mx+c$y=mx+c, $m$m is equal to the gradient which is the change in the $y$y-value for every increase in the $x$x-value by $1$1. This means we have a value of $m=0$m=0. That is, the gradient of the line is $0$0.

If we extended the table of values one place to the left, i.e. when $x=0$x=0, we would find that $y$y still has a value of $4$4, this means we have a $y$y-intercept of $4$4. This means we have a value of $c=4$c=4.

Putting it all together we end up at the equation $y=0x+4$y=0x+4 which simplifies to $y=4$y=4. 

But what if the values for $x$x and $y$y were reversed?

Consider the following table of values

We can see, in this case, that the $x$x-value is not actually changing, and the $y$y-value is increasing by $1$1 each time. Whatever the $y$y-value is, $x$x is always equal to $4$4, so the equation for this table of values is simply $x=4$x=4.

It doesn't actually matter what the increase in $y$y-value is in this case - the table could be as follows, and it would still have the same equation $x=4$x=4. 

In this case the gradient is considered to be undefined.

Here are two examples of horizontal lines:

Horizontal lines $y=2$y=2 and $y=-3$y=−3

Here are two examples of vertical lines:

Vertical lines $x=-1$x=−1 and $x=4$x=4

Practice questions

Question 1

Question 2

Question 3

Question 4