Find the area from 3 vertex coordinates — no height needed
Reviewed by [email protected], Geometry Calculator Developer & Online Math Educator Last updated May 14, 2026
When you know the coordinates of the three vertices of a triangle, you don't need to compute side lengths or height first — there is a direct formula that gives the area in one step. It is the determinant of a 3×3 matrix containing the coordinates, divided by 2. The same formula doubles as a collinearity test: if the area comes out as 0, the three points lie on a single straight line.
| Name | Formula | Notes |
|---|---|---|
| Standard Formula | A = ½ × |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| |
The three vertices are (x₁, y₁), (x₂, y₂), (x₃, y₃). Always take the absolute value; the orientation determines the sign before |·|. |
| Determinant Form | A = ½ × |det([x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1])| |
Identical to the standard formula — expand the 3×3 determinant along the third column and you get the standard form. |
| Shoelace Form (compact) | A = ½ × |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| |
Same expression rearranged. The "shoelace" name comes from the diagonal-product visualization. |
| Vector / Cross Product | A = ½ × |⃗AB × ⃗AC| |
For vertices A, B, C: half the magnitude of the cross product of two edge vectors. Works in 3D too (gives the planar area of the triangle). |
| Collinearity Test | A = 0 ⟺ 3 points are collinear |
If the formula returns 0, the three points lie on one line. Equivalent to slopes between pairs being equal. |
| Signed Area (orientation) | A_signed = ½ × [x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)] |
Without the absolute value: positive if vertices are listed counter-clockwise, negative if clockwise. Useful in computational geometry. |
| Equilateral Special Case | A = (√3/4) × s² |
When all three sides are equal (s). Verify the coordinate formula matches by computing s² = (x₂−x₁)² + (y₂−y₁)². |
Plug in your numbers and get instant step-by-step results.