Pretty much every C1 student I’ve ever worked with has said the same thing: I don’t get the big questions at the end of the paper. The ones with the curves and the equation of a line and the tangents and the turning points — what’s that all about?

So, I thought I’d put together a guide on how to think about all that stuff. Or how I think about all that stuff. It’s a three-part series: this week, I’m going to show you how to think about the equation of a line; next week, it’s all about curves, tangents and normals; and in a fortnight, I’ll extend it to circles (which might be C2, depending on your exam board).

So, what is a straight line?

It’s one of those things that you can point at and say ‘that’s a line’, but it’s a bit tricky to define a) mathematically and b) usefully. The traditional definition is ‘the shortest distance between two points, extended infinitely in both directions.’ See what I mean about usefully?

I’m going to explain it a bit more hand-wavily: it’s a set of points at a fixed angle from the $x$-axis. That’s not quite enough to define a straight line (you also need to know a point on the line), but it’s a good place to start.

As far as you’re concerned, a straight line is a mathematical object with two properties: — A gradient, which tells you how steep it is; — A point, which tells you where the line is.

It also has something you can work out from those properties: — An equation, which tells you whether a given point is on the line.

(And, of course, you can work backwards from the equation of a line to get the gradient and/or any point on it).

There are a few ways you can find out how steep a line is. You might be told the gradient explicitly; you might be given two points and need to work out the gradient between them; you might be told it’s tangent or normal to a curve (of which more next week); or you might be told it’s parallel or perpendicular to another line. I’ll take you through cases 2 and 4 right here.

The gradient between two points

The gradient just means ‘how far up does the line go every time you go across one unit?’ So, if you know the points $(-4, −3)$ and $(2,15)$ are on the line, you can say “to go from one point to the other, I go across 6 and up 18 — that means I go up 3 units (18÷6) for each unit I go across. The gradient of that line is three.

There’s a formula for it, if you prefer: the gradient, $m$, of the line connecting $(x_1, y_1)$ to $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$ — the change in $y$ divided by the change in $x$. Be careful to keep the co-ordinates the same way round!

The gradient of parallel and perpendicular lines

If you know your line is parallel to another line, you’re in luck: the two lines have the same gradient. If you work out the gradient of one, you have the gradient of the other. Boom.

It’s a bit trickier if they’re perpendicular. You need to find the negative reciprocal of the other gradient, which just means “do −1 divided by it”. If your original gradient was 2, the perpendicular gradient would be $-\frac{1}{2}$. If the original gradient was $\frac{2}{3}$, the perpendicular gradient would be $-\frac{3}{2}$.

Finding the equation of a line

The equation of a line just tells you whether a point — with coordinates $x$ and $y$ — lies on the line or not. You just work out the two sides of the equation using the values you know; if they’re the same, you’re on the line.

For example, $(2,3)$ is on the line $y = x + 1$; the left-hand side is 3 (because the $y$-value is 3) and the right-hand side is $2+1=3$. They’re the same so the point is on the line. $(-3, 4)$ isn’t on the line, because 4 isn’t the same as $-3+1=-2$. (This technique works for the equation of any curve: you just work out both sides of the equation and see if they’re equal).

There are many possible ways to write the equation of a line. Infinitely many, in fact. However, there are a couple that are generally more acceptable than others. One of them, you’ve been playing with since Year 9; it’s the old warhorse $y=mx+c$. The other, which you’ll see a lot in C1, is $ax+by+c=0$, and should probably be your default way of writing it down; that said, if they don’t tell you which way to write it down, either form is fine.

You don’t really want to work out the equation of a line using those forms, though: you want to use the Grown-Up Equation Of A Line, which looks like this:


where $m$ is the gradient of the line and $(x_0, y_0)$ is a point you’re given.

Once you know the gradient and a point on the line, you can simply throw the values into this equation and rearrange to get the equation of the line in either form.

When the gradient is a fraction

Top tip: if the gradient is a fraction — which it probably is — you can save yourself a lot of hassle by cross-multiplying the bottom of it over to the other side. This is especially useful if you’re looking for the $ax + by + c = 0$ form, since it all falls out nicely. Don’t believe me? Watch this, with $m=\frac{2}{3}$, going through $(5,7)$.

Start with the template equation: $(y-y_0) = m(x-x_0)$ Sub in the values for the point and gradient: $(y-7) = \frac{2}{3}(x-5)$ Multiply everything by 3: $3y-7 = 2(x-5)$ — already looking nicer! Multiply out: $3y-7=2x-10$ Bring it all to the right: $0 = 2x - 3y −3$

Not hard at all!