Straight line equation

You may be sitting there thinking “I know how to find the equation of a straight line! It’s just y = mx + c right?”

You would be right. But actually, there’s more than one way to express the equation of a straight line, depending on what information you have about the line - sometimes, another form is easier to use.

y = mx + c

This is the form you’re used to seeing. The values in the equation represent:

y: the y-coordinate of any point on the line, which varies based on the x-coordinate.
x: the x-coordinate of any point on the line (this goes on forever to represent all values on a graph).
m: the gradient of the line - how steep it is. This is calculated by the change in y divided by the change in x between any two points on the line.
c: the y-intercept of the line - where it crosses the y-axis (where x = 0).

y - y_1 = m(x - x_1)

You may or may not have seen this form before. It’s called point-gradient form and we use it in this situation:

We know the gradient of the line, m.
We know one point on the line, (x_1, y_1).

Remember: y and x are the coordinates of any point on the line - we do not know these values. The equation of any line/curve will have an x and y in it - don’t be caught out by trying to find the values of them.

This form is useful because we can substitute in the values we know (the gradient and the point) to get an equation that only has x and y in it - which is what we want for the equation of a line.

The values in the equation represent:

x_1: the x-coordinate of the known point on the line.
y_1: the y-coordinate of the known point on the line.
m: the gradient of the line.

Find the equation of the line with gradient 3 that passes through the point (2, 5).

Values we know:
- m = 3
- x_1 = 2
- y_1 = 5
Substitute into the formula:
- y - y_1 = m(x - x_1)
- y - 5 = 3(x - 2)
Technically, we’ve now found the equation, but we can still simplify it:
- y - 5 = 3x - 6
- y = 3x - 6 + 5
- y = 3x - 1
Answer: The equation of the line is y = 3x - 1.

Find the equation of the line with gradient -2 that passes through the point (-1, 4).

Values we know:
- m = -2
- x_1 = -1
- y_1 = 4
Substitute into the formula:
- y - y_1 = m(x - x_1)
- y - 4 = -2(x - (-1))
Again, we’ve now found the equation, but we can still simplify it:
- y - 4 = -2(x + 1)
- y - 4 = -2x - 2
- y = -2x - 2 + 4
- y = -2x + 2
Answer: The equation of the line is y = -2x + 2.

Finding the gradient from two points

There isn’t an explicit formula for finding the equation of a straight line from two points.

Technically, there could be. In fact, why don’t I make one up now?
Below, (x_1, y_1) and (x_2, y_2) are the two points we know:

y-y_1 = (\frac{y_2 - y_1}{x_2 - x_1})(x - x_1)

That isn’t too nice to look at. But actually, you may notice that all I have done is substitute the formula for gradient (m = \frac{y_2 - y_1}{x_2 - x_1}) into the point-gradient form of the straight line equation (y - y_1 = m(x - x_1)).

That’s because we can calculate the gradient from two points like this:

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

(this is the mathematical way of saying ‘change in y over change in x’).

Once we have the gradient, we can use either of the two forms of the straight line equation above to find the full equation of the line. The second form is usually easier to use, since we already have a point.

We can also pick either point to use in the equation - it doesn’t matter which one we choose - but it generally is easier to use the first point listed :)

Find the equation of the line that passes through the points (1, 2) and (4, 8).

Values we know:
- x_1 = 1
- y_1 = 2
- x_2 = 4
- y_2 = 8
First, we find the gradient:
- m = \frac{y_2 - y_1}{x_2 - x_1}
- = \frac{8 - 2}{4 - 1}
- = \frac{6}{3}
- = 2
Next, we substitute the gradient and one of the points into the point-gradient form of the equation:
- y - y_1 = m(x - x_1)
- y - 2 = 2(x - 1)
Finally, we can simplify the equation:
- y - 2 = 2x - 2
- y = 2x - 2 + 2
- y = 2x
Answer: The equation of the line is y = 2x.

Find the equation of the line that passes through the points (-2, 3) and (1, -1).

Values we know:
- x_1 = -2
- y_1 = 3
- x_2 = 1
- y_2 = -1
First, we find the gradient:
- m = \frac{y_2 - y_1}{x_2 - x_1}
- = \frac{-1 - 3}{1 - (-2)}
- = \frac{-4}{1 + 2}
- = \frac{-4}{3}
Next, we substitute the gradient and one of the points into the point-gradient form of the equation:
- y - y_1 = m(x - x_1)
- y - 3 = -\frac{4}{3}(x - (-2))
Finally, we can simplify the equation:
- y - 3 = -\frac{4}{3}(x + 2)
- y - 3 = -\frac{4}{3}x - \frac{8}{3}
- y = -\frac{4}{3}x - \frac{8}{3} + 3
- y = -\frac{4}{3}x - \frac{8}{3} + \frac{9}{3}
- y = -\frac{4}{3}x + \frac{1}{3}
Answer: The equation of the line is y = -\frac{4}{3}x + \frac{1}{3}.

flashcards

Question	Answer
What is the gradient-intercept form of a straight line equation?	y = mx + c
In y = mx + c, what does m represent?	The gradient of the line (steepness), calculated as change in y divided by change in x.
In y = mx + c, what does c represent?	The y-intercept, where the line crosses the y-axis (x = 0).
What is point-gradient form of a straight line equation?	y - y_1 = m(x - x_1)
When do you use y - y_1 = m(x - x_1)?	When you know the gradient m and one point (x_1, y_1) on the line.
In y - y_1 = m(x - x_1), what do x_1 and y_1 represent?	The x-coordinate and y-coordinate of the known point on the line.
Find the equation of the line with gradient 3 through point (2, 5).	y = 3x - 1
Find the equation of the line with gradient -2 through point (-1, 4).	y = -2x + 2
How do you find the gradient from two points (x_1, y_1) and (x_2, y_2)?	m = \frac{y_2 - y_1}{x_2 - x_1}
Find the equation of the line through (1, 2) and (4, 8).	y = 2x
Find the equation of the line through (-2, 3) and (1, -1).	y = -\frac{4}{3}x + \frac{1}{3}
What is the combined formula for the equation of a line from two points?	y - y_1 = (\frac{y_2 - y_1}{x_2 - x_1})(x - x_1)