Taking it further

A possible first step to try to generalise our findings is to take a line that does not go through the origin.

Think about what was special about a line through the origin. Why did this make it a good starting point?

If we choose a line such as \(y = x +1\) then we can draw points such that the \(x\) intercepts differ by \(2\) and the \(y\) intercepts differ by \(3\).

Plot of y = x + 1.

If we drew a line through these two points, would the two lines satisfy the required conditions?

By choosing \(y = x +1\), we cannot modify the gradient or the intercepts of this line. Instead, perhaps it would be more useful to focus on one aspect. We could pick a point that the line passes through, or we could decide on the gradient of the line.

To explore a more general approach and get a feel for what is happening you can use the GeoGebra applet below or use this Desmos graph. You can click on the circles on the left to hide and show different lines as you wish.

Why are there four other lines along with the initial black line? Do these cover all possibilities? Could there be any other lines we should consider?

Below are discussions of three different approaches.

It is perhaps easiest to visualise what the solutions will look like here compared to the two methods below. Given you know the gradients, the solutions will just be translations of the lines up and down the y axis.

Let’s start by picking a gradient of two and take our first line to be \(y = 2x + c\). Then we can take the case where the second line has a gradient of \(3\) and a \(y\) intercept of \(c + 3\).

When \(y = 0\), the \(x\) values will be \(x= -\dfrac{c}{2}\) and \(x = -\dfrac{c+3}{3}\).

The situation in the graph below is \(-\dfrac{c}{2}-\left(-\dfrac{c+3}{3}\right) = 2\).

Plots of y = 3 x plus c plus 3 and y = 2 x plus c and their intersection points with the x axis.

Subtracting the intercepts the other way \(-\dfrac{c+3}{3}-\left(-\dfrac{c}{2}\right) = 2\) would come from a graph that looks like this.

Plots of y = 3 x plus c plus 3 and y = 2 x plus c and their intersection points with the x axis.

What solutions arise from these two situations?

What sort of equations are you solving to find the solutions?

Can you find all the possible pairs of lines when the gradient is \(2\)? How many are there altogether?

If you take the lines to be \(y = mx+c\) and \(y = (m+1)x + c + 3\), can you find \(c\) in terms of \(m\)? What sort of expression is it? You should end up with two solutions.

Can you continue to find all the possible solutions for \(c\) in terms of \(m\)? You might start to notice some symmetry appearing…

Let us take our first line to be \(y= mx +1\). We can take the second line to be \(y = (m+1)x + 4\).

Remember this is only one of four possibilities that the second line could be (think of the graph above).

We have our \(y\) intercepts differing by three, we have our gradients differing by one, so we just need a difference of two in our \(x\) intercepts.

When \(y = 0\), \(x = -\dfrac{1}{m}\) in our first line and \(x = -\dfrac{4}{m+1}\) in our second.

Plots of y = m + 1 x + 4 and y = m x + 1 and their intersection points with the x axis.

We only want a ‘difference’ of \(2\) in our \(x\) intercepts, so we could subtract them either way round.

  • Will both lead to solutions?

  • How many solutions do you get?

If we continued with our line of \(y = mx +1\) then what else could the second line be?

Remember we started with \(y = (m+1)x + 4\). What can we change to find a different set of solutions?

We are used to thinking about the \(y\) intercept and the gradient when we deal with straight lines, so this may not have been an obvious place to start. However, this actually brings out the relationship between them quite nicely.

Plot of y = m x plus c.

The \(x\) intercept of this line is \(x= -\dfrac{c}{m}\), so the \(x\) intercept of the second line could be \(x=-\dfrac{c}{m}-2\) or \(x = -\dfrac{c}{m}+2\).

We have two possible \(x\) intercepts and two possible gradients (\(m+1\) and \(m-1\)). Choosing one of each should allow us to find the equation of a line.

You may find it beneficial to use the \(y - y_0 = m(x - x_0)\) form of a straight line.

You should now have a line where the gradient differs by \(1\) and the \(x\) intercept by \(2\) from the line \(y = mx+c\).

Using the fact that the \(y\) intercepts need to differ by three, can you form an equation in terms of \(c\) and \(m\)?

Can you rearrange it to give you \(c\) in terms of \(m\)?

Remember, we only chose one of our \(x\) intercepts and one of our possible gradients. So are there more equations you could find linking \(m\) and \(c\)? How many will there be altogether?