CALCULATOR · MATH · GEOMETRY
Three ways in, one answer out: area from three vertex coordinates via the shoelace formula — (0,0), (4,0), (0,3) gives 6 square units — area from base × height, or solve for the missing height or base when the area is known. Pick the mode, enter the numbers, get the answer.
Pick a mode: three vertices computes the area with the shoelace formula A = |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| ÷ 2; base × height does the classic half-product; solve height and solve base rearrange the formula when the area is the known value. Presets: the (0,0),(4,0),(0,3) default, a 3-4-5 right triangle, a base–height pair, both solver directions, and the collinear edge case.
The shoelace formula turns coordinates into area in one line. A = |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| ÷ 2. For (0,0), (4,0), (0,3): |0×(0−3) + 4×(3−0) + 0×(0−0)| ÷ 2 = 12 ÷ 2 = 6 square units. The 3-4-5 right triangle (0,0),(3,0),(0,4) gives 6 square units; (0,0),(6,0),(0,8) gives 24 square units. The solver modes invert the schoolbook formula: area 50 with base 8 needs height 12.5; area 50 with height 10 needs base 10.
Setup: vertices (0,0), (4,0), (0,3) — a right triangle with legs 4 and 3.
Line 1 — the shoelace sum. S = x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂) = 0×(0−3) + 4×(3−0) + 0×(0−0) = 12.
Line 2 — the absolute half. A = |S| ÷ 2 = 12 ÷ 2 = 6 square units. The sign of S only records whether the vertices were listed clockwise or counterclockwise — the absolute value erases it.
Line 3 — the schoolbook cross-check. Base 4, height 3: A = 4 × 3 ÷ 2 = 6 square units. Same answer, different road.
Branch — solve height (area 50, base 8): height = 2 × 50 ÷ 8 = 12.5. Branch — solve base (area 50, height 10): base = 2 × 50 ÷ 10 = 10.
Branch — collinear points (0,0),(1,1),(2,2): S = 0×(1−2) + 1×(2−0) + 2×(0−1) = 0 → no triangle. The solver refuses instead of printing a zero area, because a zero area is almost always a data-entry slip.
| Output | Formula | What it means |
|---|---|---|
| Area from vertices | |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| ÷ 2 | 6 square units at the defaults — the enclosed region, in square units of whatever units the coordinates use. |
| Area from base × height | base × height ÷ 2 | 35 square units for the 10 × 7 preset — the same area by the schoolbook route. |
| Solved height | 2 × area ÷ base | 12.5 for area 50, base 8 — the perpendicular height that reproduces the known area. |
| Solved base | 2 × area ÷ height | 10 for area 50, height 10 — the base that reproduces the known area. |
The shoelace sum is exact arithmetic on your inputs — no rounding enters until the final display, which shows up to 2 decimals. The sign convention: the raw shoelace sum is positive for counterclockwise vertex order and negative for clockwise; the formula’s absolute value makes the order irrelevant, so listing vertices in any order gives the same area. Collinear points (all three on one line) enclose zero area and are rejected as an error rather than returned as 0 — a true zero-area triangle is degenerate by definition. Coordinates beyond ±10¹² are refused to avoid floating-point garbage. The solver modes assume the height is perpendicular to the base — a slanted side is not a height. Units are whatever your inputs use (feet, meters, pixels); the area comes out in square units.
Bright Side Kit. "Triangle Area from Coordinates: Shoelace Formula Solver | Bright Side Kit." https://brightoolbox.com/area-triangle-coordinates/ Accessed October 24, 2026.Data sourcesA = |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| ÷ 2. Multiply each x by the difference of the other two y’s, add the three terms, take the absolute value, halve it. For (0,0),(4,0),(0,3): |0 + 12 + 0| ÷ 2 = 6. It’s called “shoelace” because the cross-multiplication pattern looks like laced shoes when you write the coordinates in two columns.
No. Listing the vertices clockwise gives a negative shoelace sum; counterclockwise gives a positive one — but the formula takes the absolute value, so both orders return the same area. What matters is that you use the correct three points, not the sequence.
Three points on one line enclose zero area — there is no triangle. The shoelace sum comes out to 0, which is degenerate by definition. The solver treats it as an error because in practice it’s almost always a typo — a duplicated or mis-keyed vertex — and a silent zero would hide the mistake.
Square units of whatever your inputs use. Coordinates in feet give square feet; in meters, square meters; in pixels, square pixels. The formula is unit-agnostic — it only requires all three vertices to use the same unit.
Height = 2 × area ÷ base. It’s the schoolbook formula A = base × height ÷ 2 solved for height. Area 50 with base 8 → height 12.5. Use the “solve for height” mode above — it does the rearrangement for you.
Yes, but the height must still be perpendicular to it. Any side can serve as the base; the height is the perpendicular distance from the opposite vertex to that side’s line. The most common mistake is using another side’s length as the “height” — that only works for right triangles.
Heron’s formula uses three side lengths; the shoelace formula uses three coordinates. If you know the sides, Heron’s (A = √(s(s−a)(s−b)(s−c))) avoids coordinates entirely. If you know coordinates, shoelace skips the side-length step. Same area either way.