Area of a triangle and collinearity
Find the area of a triangle from the coordinates of its three vertices using one formula. If the area is zero, the points lie on one straight line (collinear).
What you need to know
A triangle has three vertices (corners). If you know their coordinates, you can find the area without measuring any side. Name the vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃). Take them in anticlockwise order (your textbook calls it counter-clockwise: the opposite way to a clock's hands). Area is never negative. If your answer comes out negative, you took the points clockwise: just write the positive value. Points that lie on one straight line are called collinear. They make no triangle, so the area is 0.
Column method: a quick way to use the formula
- 1Plot the three points roughly and name them A, B, C going anticlockwise.
- 2Write the x-coordinates in one column and the y-coordinates in the next. Repeat the first point at the bottom.
- 3First sum: multiply each x by the y in the row below it, and add (x₁y₂ + x₂y₃ + x₃y₁).
- 4Second sum: multiply each y by the x in the row below it, and add (y₁x₂ + y₂x₃ + y₃x₁).
- 5★Area = (1/2)(first sum − second sum). Write the unit: sq. units.
★ key step
Example 1: area from three vertices
Find the area of the triangle with vertices A(−2, −3), B(4, −1) and C(2, 5).
- Formula: Area = (1/2){x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)}
- Here x₁ = −2, y₁ = −3, x₂ = 4, y₂ = −1, x₃ = 2, y₃ = 5. The points go anticlockwise.
- x₁(y₂ − y₃) = −2(−1 − 5) = −2 × (−6) = 12
- x₂(y₃ − y₁) = 4(5 − (−3)) = 4 × 8 = 32
- x₃(y₁ − y₂) = 2(−3 − (−1)) = 2 × (−2) = −4
- Area = (1/2)(12 + 32 − 4) = (1/2)(40) = 20
- Check with the column method: first sum = 2 + 20 − 6 = 16, second sum = −12 − 2 − 10 = −24, and (1/2)(16 + 24) = 20.
Area of triangle ABC = 20 sq. units
Example 2: three places on one straight road
On a village map, a well is at W(1, 4), a temple at T(3, k) and a bus stop at S(7, −2). A straight road passes through all three. Find k.
- Places on one straight road are collinear, so the area of triangle WTS = 0.
- (1/2){x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)} = 0
- Multiply by 2: 1(k − (−2)) + 3(−2 − 4) + 7(4 − k) = 0
- k + 2 − 18 + 28 − 7k = 0
- 12 − 6k = 0, so k = 2
- Check: slope WT = (2 − 4)/(3 − 1) = −1 and slope TS = (−2 − 2)/(7 − 3) = −1. The slopes match.
k = 2, so the temple is at T(3, 2).
Land surveyors in Tamil Nadu mark the corners of a field as points on a map. The coordinate formula then gives the area of the field without walking along every side.
2 marks: “Show that the points (2, 5), (4, 6) and (8, 8) are collinear.” Area = (1/2){2(6 − 8) + 4(8 − 5) + 8(5 − 6)} = (1/2)(−4 + 12 − 8) = 0. The area is zero, so the points are **collinear**.
5 marks: “The vertices A(−2, 1), B(k, −3) and C(5, 3), taken in order (anticlockwise), form a triangle of area 20 sq. units. Find k.” (1/2){−2(−3 − 3) + k(3 − 1) + 5(1 + 3)} = (1/2)(32 + 2k) = 16 + k. Taken anticlockwise, this value is positive, so 16 + k = 20 and **k = 4**. (If no order is given, 16 + k = −20 also works, giving k = −36.)
5 marks: “Find the area of the triangle whose vertices are A(2, −2), B(−3, 5) and C(−2, −4).” (1/2){2(5 + 4) + (−3)(−4 + 2) + (−2)(−2 − 5)} = (1/2)(18 + 6 + 14) = **19 sq. units**.
Zero area, straight line: three friends standing in a straight queue make no triangle.