Area of a triangle from its vertices
A farmer knows the corner positions of a triangular plot on a map grid but not its sides. Using only the corner coordinates, a determinant gives the area directly.
The area of the triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is half the absolute value of the determinant of the rows (x, y, 1). Area is never negative, so take the positive value.
Formula: Δ = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
Rule 1: Use the formula — Put the three vertices in rows with a column of 1s.
- Easy: (0, 0), (4, 0), (0, 3): Δ = ½ |4 × 3| = 6.
- Medium: (1, 1), (5, 1), (3, 4): determinant = −3 − 2 + 17 = 12, so Δ = 6.
- Exam-type: (−1, 2), (3, −2), (5, 4): 6 + 6 + 20 = 32, so Δ = 16.
Rule 2: Work backwards — Form an equation in the unknown coordinate.
- Easy: (0, 0), (4, 0), (2, k) has area ½ × 4 × |k| = 6, so k = 3 or −3.
- Medium: A triangle with vertices (0, 0), (8, 0), (3, −6) has area ½ × 8 × 6 = 24.
- Exam-type: The area is 0 exactly when the three points are on one line.
Use absolute value at the end: area cannot be negative.
Leaving a negative area.
Half the determinant of [x y 1], then take the positive value.
Practice questions
- 1Find the area of the triangle with vertices (0, 0), (6, 0), (0, 4).
- 2Find the area of the triangle with vertices (2, 0), (6, 0), (4, 5).
- 3Find the area of the triangle with vertices (1, 1), (4, 1), (2, 5).
- 4Find the area of the triangle with vertices (−2, −1), (2, 3), (0, 5).
- 5Find the area of the triangle with vertices (3, 2), (−1, 4), (1, −2).
- 6Find k if the triangle (0, 0), (4, 0), (2, k) has area 8 square units.
- 7Find the area of the triangle with vertices (1, 2), (3, 6), (5, 10). What does the answer tell you?
- 8Find the area of the triangle with vertices (0, 0), (8, 0), (3, −6).
Show answers (try them yourself first)
- 1½ × 6 × 4 = 12
- 2½ |2(0 − 5) + 6(5 − 0) + 4(0 − 0)| = 10
- 3½ |1(1 − 5) + 4(5 − 1) + 2(1 − 1)| = 6
- 4½ |−2(3 − 5) + 2(5 + 1) + 0| = 8
- 5½ |3(4 + 2) − 1(−2 − 2) + 1(2 − 4)| = 10
- 6½ × 4 × |k| = 8, so k = 4 or −4
- 7The area is 0, so the three points lie on one straight line
- 824 (the value −24 becomes 24 after the absolute value)
Collinear points and equation of a line
Three milestones along a straight road make a triangle of zero area. This fact turns a question about points on a line into a determinant that equals zero.
Three points are collinear exactly when the area of the triangle they form is 0, i.e. the determinant with rows (x, y, 1) is 0. The line through (x₁, y₁) and (x₂, y₂) is the set of points (x, y) with |x y 1 ; x₁ y₁ 1 ; x₂ y₂ 1| = 0.
Formula: Collinear ⇔ |x₁ y₁ 1 ; x₂ y₂ 1 ; x₃ y₃ 1| = 0
Rule 1: Test or complete collinearity — Set the determinant to zero.
- Easy: (1, 1), (2, 2), (3, 3) are collinear because the determinant is 0.
- Medium: (1, 2), (2, 4), (3, k) are collinear when k − 6 = 0, so k = 6.
- Exam-type: (1, k), (2, 4), (3, 7) are collinear when k − 1 = 0, so k = 1.
Rule 2: Equation of a line from two points — Expand |x y 1 ; x₁ y₁ 1 ; x₂ y₂ 1| = 0.
- Easy: The line through (1, 2) and (3, 6) is −4x + 2y = 0, i.e. y = 2x.
- Medium: The line through (0, 1) and (2, 5) is y = 2x + 1.
- Exam-type: The line through (1, 1) and (4, 7) is y = 2x − 1.
The determinant method needs no slope: it also works for vertical lines.
Forgetting that a zero determinant means collinear, not "no triangle possible by error".
Zero area = same line.
Practice questions
- 1Are (1, 1), (2, 3), (3, 5) collinear?
- 2Are (0, 0), (1, 2), (2, 5) collinear?
- 3Find k so that (1, 2), (2, 4), (3, k) are collinear.
- 4Find k so that (2, 3), (4, 7), (6, k) are collinear.
- 5Find the equation of the line through (1, 2) and (3, 6) using a determinant.
- 6Find the line through (0, 1) and (2, 5).
- 7Find the line through (1, 1) and (4, 7).
- 8For which k are (1, k), (2, 4), (3, 7) collinear?
Show answers (try them yourself first)
- 1Yes, the determinant is −2 + 1 + 1 = 0
- 2No, the determinant is 1, not 0
- 3k = 6
- 4The slope between the first two points is 2, so k = 11
- 5−4x + 2y = 0, i.e. y = 2x
- 6−4x + 2y − 2 = 0, i.e. y = 2x + 1
- 7−6x + 3y + 3 = 0, i.e. y = 2x − 1
- 8k − 1 = 0, so k = 1