The Line Through Two Points

You have a formula that turns two points into a gradient, and a formula that turns a gradient and a point into an equation. What happens if you run them one after the other?

Step one: use the gradient formula on the two points. Step two: put that gradient and either point into point–gradient form. Step three: substitute the other point into your answer, which checks the whole thing for free.

Nothing in this lesson is new. Lesson 2 turned two points into a gradient. Lesson 6 turned a gradient and a point into an equation. Running them in that order turns two points into an equation.

For $A(-2, 9)$ and $B(4, -3)$:

$m = \dfrac{-3 - 9}{4 - (-2)} = \dfrac{-12}{6} = -2$

Then, using $A$: $y - 9 = -2(x + 2)$, which gives $y - 9 = -2x - 4$ and $y = -2x + 5$.

The order is forced. Point–gradient form cannot be used until a gradient exists, so the gradient step is not optional and cannot be skipped by substituting harder.

Repeat the second step with $B(4, -3)$ instead:

$y - (-3) = -2(x - 4)$, so $y + 3 = -2x + 8$ and $y = -2x + 5$.

The same equation, as it must be. Both points lie on the line, and the line has one equation.

So choose whichever point makes the arithmetic easier. A point with positive coordinates avoids double negatives; a point with a zero in it removes a term entirely; a point that happens to be the y-intercept lets you write the answer down immediately.

This is a real choice with a real saving, and it costs nothing, because the answer is not affected by which one you pick.

Because only one point was needed for the second step, the other is still unused, and that makes it a proper check.

Having found $y = -2x + 5$ using $A$, substitute $B(4, -3)$: the right-hand side gives $-2(4) + 5 = -3$, which matches the y-coordinate of $B$. The line passes through both points, so it is correct.

Compare this with re-reading your own working. Re-reading catches only the errors you notice, and you tend not to notice the same slip twice. Substituting an unused point tests the answer against information the working has not yet touched, so it can genuinely disagree.

Do it every time. It costs one line and catches sign errors, expansion errors and gradient errors all at once.

Look at the two points before starting, and two cases will announce themselves.

Equal y-coordinates, such as $(-5, 4)$ and $(3, 4)$. The gradient is zero and the line is horizontal, so the equation is $y = 4$. The general method also gets there, but there is no need to run it.

Equal x-coordinates, such as $(6, 1)$ and $(6, -7)$. The gradient is undefined and the method breaks at the first step, since the denominator is zero. The line is vertical, so the equation is $x = 6$.

The second case is the one that matters, because a student who has not looked first will divide by zero and may write down whatever the calculator says. Glance at the coordinates before you substitute. If the x-coordinates match, stop and write the vertical equation.

Many practical questions give two matched pairs of values rather than two labelled points, but the mathematics is identical.

A printer charges $\$70$ for $100$ flyers and $\$130$ for $300$ flyers, and the cost is known to be linear. Treating quantity as $x$ and cost as $y$ gives the points $(100, 70)$ and $(300, 130)$.

The gradient is $\dfrac{130 - 70}{300 - 100} = \dfrac{60}{200} = 0.3$, so each extra flyer costs $30$ cents.

Using $(100, 70)$: $y - 70 = 0.3(x - 100)$, so $y = 0.3x + 40$. The constant $40$ is the fixed setup charge, the cost of printing zero flyers.

The gradient and the y-intercept both have meanings here, and reporting them in words is usually part of the question. A rate per unit and a fixed charge, respectively, is the pattern to expect.

Five items. Look at the coordinates first, in case the line is horizontal or vertical.

1. 1 Find the equation of the line through $(0, 3)$ and $(2, 7)$.

   Show answer
   $m = 2$, and $(0, 3)$ is the y-intercept.$y = 2x + 3$

2. 2 Find the equation of the line through $(1, 5)$ and $(3, 1)$.

   Show answer
   $m = \dfrac{1 - 5}{3 - 1} = -2$, then use $(1, 5)$.$y = -2x + 7$

3. 3 Find the equation of the line through $(-4, 2)$ and $(6, 2)$.

   Show answer
   The y-coordinates match, so the line is horizontal.$y = 2$

4. 4 Find the equation of the line through $(5, -1)$ and $(5, 8)$.

   Show answer
   The x-coordinates match, so the gradient is undefined and the line is vertical.$x = 5$

5. 5 Find the equation of the line through $(-3, -1)$ and $(3, 3)$, in gradient–intercept form.

   Show answer
   $m = \dfrac{4}{6} = \dfrac{2}{3}$, then use $(3, 3)$.$y = \tfrac{2}{3}x + 1$