Case 1
- x₁
- 0
- y₁
- 0
- x₂
- 4
- y₂
- 0
- x₃
- 0
- y₃
- 3
6.000000
Open with these valuesTriangle Area from Coordinates
6.000000units²
Result: 6.000000 units²Three points are enough — no base, no height, no angles. The shoelace formula crosses the coordinates and halves what is left. All six numbers share one length unit and the area comes back squared in it. Collinear points return zero, which is the honest answer.
The numbers at a glance
Held fixed: x₁ 0.000, y₁ 0.000, x₂ 4.000, y₂ 0.000, x₃ 0.000.
| y₃ | Result |
|---|---|
| 0.000 | 0.000000 |
| 1.000 | 2.000000 |
| 2.000 | 4.000000 |
| 3.000Your value | 6.000000 |
| 4.000 | 8.000000 |
| 5.000 | 10.000000 |
| 6.000 | 12.000000 |
Worked examples
Case 2
- x₁
- -1
- y₁
- -1
- x₂
- 3
- y₂
- -1
- x₃
- 1
- y₃
- 2
6.000000
Open with these valuesCase 3
- x₁
- 0
- y₁
- 0
- x₂
- 6
- y₂
- 0
- x₃
- 0
- y₃
- 8
24.000000
Open with these valuesHow it's calculated
A = |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| ÷ 2
- StepRead the three vertices off your drawing or map.
- StepEnter each pair as x and y, in any order of vertices.
- ResultRead the area in the square of the coordinate unit.
Reference table
| x₁, y₁, x₂, y₂, x₃, y₃ | Base × height ÷ 2 | Area |
|---|---|---|
| 0, 0, 1, 1, 2, 2 | collinear | 0.000000 |
| 0, 0, 1, 0, 0, 1 | 1 × 1 ÷ 2 | 0.500000 |
| 0.5, 0, 2.5, 0, 0, 3.5 | 2 × 3.5 ÷ 2 | 3.500000 |
| 0, 0, 4, 0, 0, 3 | 4 × 3 ÷ 2 | 6.000000 |
| -1, -1, 3, -1, 1, 2 | 4 × 3 ÷ 2 | 6.000000 |
| 0, 0, 6, 0, 0, 8 | 6 × 8 ÷ 2 | 24.000000 |
Questions
How do you find the area of a triangle from coordinates?
Cross-multiply the vertices as x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂), take the absolute value and halve it. The vertices (0,0), (4,0) and (0,3) span 6 square units.
Does the order of the vertices matter?
No. Going the other way round only flips the sign, and the absolute value removes it.
Why does the calculator return zero?
Because the three points lie on one straight line. A degenerate triangle has no area, and zero is the correct result rather than an error.
Can I use negative coordinates?
Yes. Points in any quadrant work, and the result is never negative.
Sources and last check
Information, not professional advice.