Finding the Equation of a Line

A point and a direction fix a line

Think about what it takes to nail down a single straight line. One point is not enough: infinitely many lines pass through any given point, fanning out in every direction like the spokes of a wheel. A direction alone is not enough either: all the lines with a given slope tilt the same way, a whole family of them sliding across the plane. But pin both at once, one point the line must pass through and one slope it must have, and exactly one line survives. It has to go through that point, and from there only one tilt is allowed, so there is nothing left to choose.

Both halves of that argument are things you can do to the figure below. The figure sets a line’s slope and its crossing point separately, so either one can be held while the other moves.

Fix the crossing at $3$ and leave it alone while you walk the rise from $-6$ up to $6$. Every line you produce passes through that same point on the vertical axis. The lines fan out through that point exactly as described: one point does not fix a line. Now set the rise back to $2$ over a run of $3$ and walk the crossing instead, from $-6$ up to $6$. This time the tilt never changes and the whole family slides up the plane, each line parallel to the last: a slope does not fix a line either. Only when you stop moving both does a single line survive. One caveat, since the figure is narrower than the lesson: its point is always the crossing point on the vertical axis. The form in the next section, by contrast, accepts any point at all.

Two points pin a line for the same reason, with one exception to watch for: if the two points share an $x$-coordinate, they sit on a vertical line, and a vertical line has no slope to compute (you saw why in the previous lesson). Otherwise, two points hand you the slope through the previous lesson’s formula, and once you have a slope and a point you are back to the first case. So every “find the equation” task in this lesson is really the point-and-slope task, or the vertical-line case below, sometimes with a quick slope calculation in front.

What does “the equation of the line” mean here? From the graphing lesson, the equation of a line is the rule its points obey. A pair $(x, y)$ satisfies that equation exactly when the point sits on the line. The job now is to turn “passes through this point with this slope” into such a rule.

Point-slope form

Start with the picture below. The line passes through $(1, 2)$ with slope $2$. Moving from $(1, 2)$ to any other point $(x, y)$ on the line changes $x$ by $x - 1$ and changes $y$ by $y - 2$, and a slope of $2$ means the vertical change is always twice the horizontal change:

That pattern works for any point and any slope, not only this one. Suppose you know one point on a line and the line’s slope. Call the known point $(x_1, y_1)$ and the slope $m$. To describe every point on the line at once, give it a name: let $(x, y)$ stand for an arbitrary point on the line. The slope between the known point and this general point is something you can write down with the slope formula. Because a line has only one slope, that expression has to equal the slope $m$. Setting it equal to $m$ and clearing the fraction produces the equation.

The name tells you how to read it. The two numbers $x_1$ and $y_1$ come from the point, and $m$ is the slope. To use it, substitute your point and your slope, and stop. You do not have to simplify: $y - y_1 = m(x - x_1)$ is already a complete, correct equation of the line.

Finding a line through two points

When you are handed two points instead of a point and a slope, first check whether they share an $x$-coordinate. If they do, skip straight to the vertical-line case in the next section: the equation is $x = c$, no slope needed. Otherwise, find the slope first, then feed that slope and either point into point-slope form. The phrase “either point” is worth pausing on. Both points sit on the same line, so both must produce the same line, and they do. Choosing one point over the other changes only how the equation looks before you simplify, never which line it describes.

Standard form

Point-slope form is the fastest to write, but answers are often requested in standard form, which lines the variables up on one side:

Here $A$, $B$, and $C$ are integers, and $A$ and $B$ are not both zero. By convention $A$ is zero or positive, and the three numbers share no common factor other than $1$. Standard form hides the slope and the point from view, but it treats $x$ and $y$ evenly. Standard form is also the natural way to write a vertical line, and it makes lines easy to compare and combine later.

To convert, start from point-slope form, distribute the slope, and move the $x$-term and the constants to the sides that match $Ax + By = C$. When the slope is a fraction, multiply through to clear the denominator so the coefficients come out as integers.

Horizontal and vertical lines

Two kinds of lines need a moment of care, because one of them breaks point-slope form.

A horizontal line has slope $0$. Point-slope form still works: through a point $(x_1, y_1)$ with $m = 0$ it gives $y - y_1 = 0(x - x_1)$, and the right side is just $0$, so the equation collapses to $y = y_1$. That matches what you already know, that a horizontal line has the constant equation $y = c$, where $c$ is the shared height of all its points.

A vertical line is the exception. Its slope is undefined, so there is no $m$ to substitute and point-slope form cannot be used at all. You do not need it. Every point on a vertical line has the same $x$-coordinate, so the equation simply states that: $x = c$, where $c$ is that shared $x$-value. Reach for this directly whenever the points share an $x$-coordinate, or the line is described as vertical.

Checking your equation

Every equation you write can be checked, but what a substitution actually proves depends on what you started from.

For a two-point problem, substitute both points. The point you built the equation from always checks out, whatever slope you used, so the other point does the real work: with two different $x$-coordinates, only the correct slope can make the equation true at both points.

For a point-and-slope problem, substituting the one given point is not enough by itself. Point-slope form is built so that the given point always satisfies it, whatever slope you used, so that substitution alone cannot catch a wrong slope. Use the slope to find a second point on the line, and check that one too.

Take the standard-form answer $2x + 3y = 11$ from earlier, built from the point $(4, 1)$ and slope $-\tfrac{2}{3}$. Substituting the point confirms the location:

That matches, so $(4, 1)$ is on the line. To confirm the slope too, step one run of $3$ and rise of $-2$ from $(4, 1)$ to the point $(7, -1)$, and check it as well:

Both points check out, so the location and the slope are both correct.

Common mistakes