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 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:

$(y-y\_0)=m(x-x\_0)$

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!