How steeply a line tilts, written five honest ways.
A straight line is the simplest shape there is, and yet it has many faces. Tilt your pencil up off the horizontal by some angle and you have fixed everything about a line's direction — every line through every point with that tilt is a copy of the same idea. The single number that captures that tilt is the slope, and the slope is just the tangent of the tilt angle: k = tan α. Once we have a slope and one point the line is pinned down completely, and we can write it five different ways — point-slope, slope-intercept, two-point, intercept, and the catch-all general form — each handy for a different job. This lesson shows that they are all the same line wearing different clothes, and why the last form is the only one honest enough to hold a vertical line.
Stand a line on the x-axis and ask: by how much does it tip up? Measure the angle from the positive x-axis, turning counter-clockwise, until you reach the line. That angle is the angle of inclination α. By convention we take it in the range 0° ≤ α < 180°: a horizontal line has α = 0°, a line tipping gently up to the right has a small α, a vertical line stands at α = 90°, and a line falling to the right has an obtuse α between 90° and 180°.
The inclination α describes a line's pose — which way it leans — without naming a single point on it. Two parallel lines, no matter where they sit, share the same α. That makes α the natural starting point: capture the pose first, then anchor it with a point.
The angle α is geometric; for algebra we want a number. Step one unit to the right and watch how far the line rises: that rise is the slope k. Because the run is 1 and the rise is k, the little right triangle says tan α = k / 1, so
k = tan α = y2 − y1x2 − x1 (for any two points on the line, x1 ≠ x2).
The sign of k reads off the pose directly. An acute α < 90° gives k > 0 — the line climbs left to right. An obtuse α > 90° gives k < 0 — it falls. A horizontal line has α = 0° and k = 0. And the vertical line, at α = 90°, has no slope at all: the run is 0 and dividing by 0 is forbidden, so tan 90° is undefined.
The line through A(1, 1) and B(4, 5) has slope
k = (5 − 1) / (4 − 1) = 4/3 ≈ 1.33, so α = arctan(4/3) ≈ 53.13° — an acute lean, a line that climbs.
Slope is rise over run, not run over rise — keep the change in y on top. And a vertical line has no slope; do not write k = ∞ as though infinity were a number. Describe a vertical line by its equation x = c instead.
Now anchor the pose. If a line has slope k and we know it passes through one point (x0, y0), then for any other point (x, y) on it the slope between the two must equal k: (y − y0)/(x − x0) = k. Clear the fraction and you have the point-slope form:
y − y0 = k(x − x0)
Choose the special point where the line crosses the y-axis, (0, b), and the form simplifies to the one everyone meets first, the slope-intercept form y = kx + b. Here b is the y-intercept — the height at which the line cuts the y-axis — and k is still the slope.
A line with slope k = 2 through (3, 1): point-slope gives y − 1 = 2(x − 3). Expand: y = 2x − 6 + 1 = 2x − 5, so the y-intercept is b = −5.
Sometimes you are handed two points, not a slope. No problem — compute the slope first, k = (y2 − y1)/(x2 − x1), then drop it into point-slope through either point. Written all at once this is the two-point form:
y − y1y2 − y1 = x − x1x2 − x1
A second special pair of points are the line's footprints on the axes: it crosses the x-axis at (a, 0) and the y-axis at (0, b). Those two intercepts give the elegant intercept form:
xa + yb = 1
The intercept form needs both intercepts to exist and be non-zero. It fails for a line through the origin (then a = b = 0), and for a line parallel to an axis (one intercept is missing). For those, fall back on a slope form or — best of all — the general form below.
Each form above has a blind spot, and they all share one: none of them can write a vertical line. The slope forms divide by the run; the intercept form needs two finite intercepts; the two-point form divides by x2 − x1. Multiply any of them out, move everything to one side, and you land on the one form with no blind spot at all — the general form:
Ax + By + C = 0 (A and B not both zero).
When B ≠ 0 we can solve for y and recover the slope: k = −A/B. When B = 0 the equation is Ax + C = 0, i.e. x = −C/A — a perfectly good vertical line, the one the slope forms had to refuse. That is why the rest of analytic geometry — parallel and perpendicular tests, the point-to-line distance formula — is built on this form: it never has to make an exception.
Put y = 2x − 5 into general form: move all to one side, 2x − y − 5 = 0, so A = 2, B = −1, C = −5. Check the slope: −A/B = −2/(−1) = 2 ✓. The vertical line through (3, 0) is simply x − 3 = 0 — no slope needed.
The five forms are one line in five outfits. Use point-slope when you have a point and a slope, slope-intercept to graph or read off k and b at a glance, two-point when handed two points, intercept when you know where it cuts the axes, and the general form whenever you want a single equation that never chokes — vertical lines included.
| Form | Equation | Best when… |
|---|---|---|
| Point-slope | y − y0 = k(x − x0) | you have one point + the slope |
| Slope-intercept | y = kx + b | graphing; reading k and b at a glance |
| Two-point | (y−y1)/(y2−y1) = (x−x1)/(x2−x1) | you are handed two points |
| Intercept | x/a + y/b = 1 | you know both axis crossings (a, b ≠ 0) |
| General | Ax + By + C = 0 | always — carries vertical lines; k = −A/B |
And the bridge between geometry and algebra: k = tan α = (y2 − y1)/(x2 − x1), with α = 0° → k = 0, acute α → k > 0, obtuse α → k < 0, and α = 90° → no slope (use x = c).
k = (4 − 1)/(4 − (−2)) = 3/6 = 1/2. So α = arctan(1/2) ≈ 26.57°, an acute lean — the line climbs.
Point-slope: y − 5 = −3(x − 2). Expand: y = −3x + 6 + 5 = −3x + 11 (slope-intercept). General: 3x + y − 11 = 0.
Intercept form: x/5 + y/(−2) = 1, i.e. x/5 − y/2 = 1. Slope k = (0 − (−2))/(5 − 0) = 2/5. (Or: rewrite as 2x − 5y = 10, so k = −A/B = −2/(−5) = 2/5.)
Solve for y: 2y = 4x + 6, so y = 2x + 3. Slope k = 2, y-intercept b = 3. (Check: k = −A/B = −4/(−2) = 2 ✓.)
It is vertical: α = 90° and tan 90° is undefined, so it has no slope k. The slope-intercept form requires a number k, which does not exist here. In general form it is x − 4 = 0 (A = 1, B = 0, C = −4) — the form handles it with no trouble.
k = tan 120° = −√3 ≈ −1.732 (obtuse α → negative slope, the line falls). Through the origin: y = −√3·x, i.e. √3·x + y = 0 in general form.
Six questions to lock it in. Tap the answer you think is right.
This lesson covers the high-school standards on lines in the coordinate plane. HSG-GPE.B.5 — prove the slope criteria for parallel and perpendicular lines and use them (the slope k = tan α and its sign are the groundwork here, finished in 30.3). HSA-REI.D.10 — understand that the graph of an equation in two variables is the set of all its solutions, a line. HSA-CED.A.2 — create equations in two variables to represent relationships, choosing among point-slope, slope-intercept, two-point, intercept, and the general form Ax + By + C = 0. Emphasize that all five describe the same line, and that the general form is the only one that carries vertical lines (no slope) without exception — that is why later distance and angle formulas are built on it.