Finding the equation of a line

How do I figure out the equation of a line?

There are a number of different forms of line equation. The easiest one to use depends on the information you have about the line.

Let’s, first of all, look at the form of line equation that, generally, people are most familiar with. It looks like this:

$\boldsymbol{y=mx+c}$

The $\boldsymbol{m}$ represents the gradient of the line and the $\boldsymbol{c}$ represents the y-intercept.

Example: $\boldsymbol{y=3x+7}$ is the equation of the line that has a gradient of 3 and the y-intercept at (0,7).

So if you know the gradient and the y-intercept of a line you can simply plug the numbers into this form, and you’re done.

Ok. But what if I don’t know the y-intercept?

Let’s look at another form of the line equation. This is one that most folks are a little less familiar with and it does look a bit messier at first glance. But don’t stress, we’ll figure it out. It looks like this:

$\boldsymbol{y-y_1=m(x-x_1)}$

Now, we still have $\boldsymbol{m}$ and it’s still the gradient of the line, but instead of $\boldsymbol{c}$ we have that $\boldsymbol{x_1}$ and $\boldsymbol{y_1}$ . Together $\boldsymbol{(x_1,y_1)}$ represents the co-ordinate of any one point on the line.

Example: $\boldsymbol{y-1=3(x+2)}$ is the equation of the line that has a gradient of 3 and passes through the point (-2,1).

So we no longer need to know the y-intercept. We only need to know the gradient and any one point on the line.

Note: Be careful with those negatives. In the example line equation above, notice that $\boldsymbol{x_1=-2}$ , so:

$\boldsymbol{x-x_1=x-(-2)=x+2}$

Ok. But what if I don’t know the gradient?

If you know the co-ordinates of any two points on the line, then you can figure out the gradient of the line. The gradient describes the steepness of the line. This is sometimes referred to as ‘rise over run’. How far the line rises (or falls) compared to how far it runs along from left to right. So the gradient of the line will be:

$\large{\boldsymbol{m=\frac{\text{rise}}{\text{run}}=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}}}$

where $\boldsymbol{(x_1,y_1)}$ and $\boldsymbol{(x_2,y_2)}$ are any two points on the line.

That triangle symbol is the Greek letter delta (capital Delta in fact) and stands for ‘change’. So $\boldsymbol{\Delta y}$ means ‘change in y’.

Example: If (-2,1) and (3,16) are two points on the line, then the gradient of the line is:

$\large{\boldsymbol{m=\frac{y_2-y_1}{x_2-x_1}=\frac{16-1}{3-(-2)}=\frac{15}{5}=3}}$

I’ve decided that (-2,1) is my $\boldsymbol{(x_1,y_1)}$ , and (3,16) is my $\boldsymbol{(x_2,y_2)}$ . You could do it the other way around and it would still work. You’ll get the same answer (try it and see). Just make sure to make a decision and stick to it, otherwise you’ll end up with numbers all over the place.

Now that we know the gradient, we can use it to build the line equation in either of the two forms we’ve already seen. If the y-intercept is one of the points you know, then you can now write the equation in the form $\boldsymbol{y=mx+c}$ , otherwise you can write the equation in the form $\boldsymbol{y-y_1=m(x-x_1)}$ using either of the two points you know.

Ok. But what if I don’t know any points on the line?

Find some

If you’re looking at a graph then often the easiest ones to see will be the x and y-intercepts, but you can use any points where you can clearly see the co-ordinates.

That’s all for now.

Exercise: Incidentally, the two example equations I used above do, in fact, represent the same line. See if you can do the algebra to rearrange this:

$\boldsymbol{y-1=3(x+2)}$

so that it looks like this:

$\boldsymbol{y=3x+7}$

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