The Equation of a Line Through Two Points

Two points fix a line

Through a single point there are lines of every steepness. Through two points there is only one straight line. So the two points (1, 3) and (3, 7) are enough to find its equation, y = mx + c: the gradient m first, then the y-intercept c.

The gradient first

From (1, 3) to (3, 7), the rise is 7 − 3 = 4 and the run is 3 − 1 = 2. The gradient is the rise divided by the run: $m=4/2=2$m = 4 / 2 = 2.

The one line through (1, 3) and (3, 7). The run between them is 2 and the rise is 4, so the gradient is 2. The line crosses the y-axis at 1.

Then either point fixes c

Now the line is y = 2x + c, and c is the only unknown. Every point on the line makes the equation true, so put in the coordinates of (1, 3): 3 = 2 × 1 + c, so 3 = 2 + c and c = 1. The equation is y = 2x + 1.

Either point gives the same c. With (3, 7): 7 = 2 × 3 + c = 6 + c, so again c = 1.

Point-slope form reaches the same line. Through (1, 3) with gradient 2, it is y − 3 = 2(x − 1). Expand the bracket: y − 3 = 2x − 2. Add 3 to both sides: y = 2x + 1.

Check with the other point

If (1, 3) was used to find c, then (3, 7) has not been used yet, and it makes a check. Put x = 3 into y = 2x + 1: 2 × 3 + 1 = 7, which is the y-coordinate of (3, 7). The line passes through both points, so the equation is right.

A check that fails means a slip somewhere: most often a sign in the gradient, or c found by adding where it should have been subtracted.

A line that slopes down

Find the line through (−2, 5) and (4, 2). The rise is 2 − 5 = −3 and the run is 4 − (−2) = 6, so $m=-3/6=-\frac{1}{2}$m = −3 / 6 = −1/2.

So $y=-\frac{x}{2}+c$y = −x/2 + c. Put in (−2, 5): $5=-\frac{\left(-2\right)}{2}+c=1+c$5 = −(−2)/2 + c = 1 + c, so c = 4, and the line is $y=-\frac{x}{2}+4$y = −x/2 + 4. Check with (4, 2): $-\frac{4}{2}+4=-2+4=2$−4/2 + 4 = −2 + 4 = 2.

To clear the fraction, multiply every term by 2: 2y = −x + 8. Add x to both sides: x + 2y = 8, the same line in the form Ax + By = C.

The line through (−2, 5) and (4, 2): a run of 6 with a rise of −3, so the gradient is $-\frac{1}{2}$−1/2. It crosses the y-axis at 4 and the x-axis at 8, where x + 2y = 8 with y = 0.

Same x, or same y

Through (−1, 4) and (5, 4) the rise is 0, so m = 0 and y = 0x + c. Every point on the line has y = 4, so the equation is y = 4, a level line.

Through (3, 1) and (3, 6) the run is 0, and 5 ÷ 0 has no answer: the line is vertical and has no gradient. Every point on it has x = 3, so its equation is x = 3. It cannot be written as y = mx + c, but it is Ax + By = C with A = 1, B = 0 and C = 3.

Is a third point on the line?

The check works for any point. Is (6, 1) on $y=-\frac{x}{2}+4$y = −x/2 + 4? At x = 6 the line has $y=-\frac{6}{2}+4=-3+4=1$y = −6/2 + 4 = −3 + 4 = 1, which is the y-coordinate of the point, so yes.

Is (2, 4) on it? At x = 2 the line has $y=-\frac{2}{2}+4=-1+4=3$y = −2/2 + 4 = −1 + 4 = 3, but the point has y = 4. So (2, 4) is not on the line: it is 4 − 3 = 1 unit directly above it.

The usual mistakes

Taking the point’s y-coordinate as c. For (1, 3), c is not 3: at x = 1 the 2x term is 2, and it has to be taken away, so c = 3 − 2 = 1. A point’s y-coordinate is c only when its x-coordinate is 0.

Getting the sign of c wrong. From 3 = 2 + c, c = 3 − 2 = 1, and the equation is y = 2x + 1, not y = 2x − 1.

Looking for another point. Once m is known, one point is enough to find c; a third point, such as the midpoint, only gives the same c again.