5.06 Finding the equation of a straight line

To find c we can just look at where the line crosses the y-axis. The value of y at this point is our y-intercept.

To find the gradient, we want to choose two points on the line that we can easily identify the co-ordinates of, ideally points with integer co-ordinates. Using these two points we can calculate the  gradient  , by identifying the rise and run of the line, or by using the gradient formula.

Consider the following graph. How can we work out its equation?

We could have chosen any two points on this line, but sometimes the coordinates might not be clear if they are not integer values. In this case, the point that is one unit along the x-axis from the point (0,-6) has coordinates of (1,-3) which confirms the gradient is 3 as expected.

Examples

Sometimes we don't know the value of the y-intercept, but if we know the gradient and the coordinates of one point, we can still find the equation of the line. If we don't know the gradient, but know the coordinates of two points, we can first find the gradient and then use the point-gradient formula.

Let's say we know that the gradient of the line is -2. We also know a point on the line, (2,-8).

Now, apart from this point there are infinitely many other points on this line, and we will let (x,y) represent each of them.

Well, since (x,y) and (2,-8) are points on the line, then the gradient between them will be -2.

We know that to find the gradient given two points, we use:

m =\dfrac{y_2-y_1}{x_2-x_1}

Let's apply the gradient formula to (x,y) and (2,-8):

m =\dfrac{y-(-8)}{x-2}

But we know that the gradient of the line is -2. So:

\dfrac{y-(-8)}{x-2}=-2

Rearranging this slightly, we get:

y-(-8)=-2(x-2)

You may be thinking that we should simplify the equation, and of course if you do you should get y=-2x-4. What we want to do though, is generalise our steps so that we can apply it to any case where we're given a gradient m, and a point on the line (x_1,y_1).

In the example above, the point on the line was (2,-8). Let's generalise and replace it with (x_1,y_1).

We were also given the gradient -2. Let's generalise and replace it with m.

So from y-(-8)=-2(x-2) we get: y-y_1=m\left(x-x_1\right).

We call this the point gradient formula, because when we know a point and the gradient using this rule we can easily get the equation of that line.

Given a point on the line (x_1,y_1) and the gradient m, the equation of the line is: y-y_1=m\left(x-x_1\right)

Examples