Area of a triangle using determinants
Area of a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is (1/2)|Δ|, where Δ has rows (x, y, 1). Δ = 0 means the points are collinear.
Why the formula works
In coordinate geometry you met area = (1/2)[x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)]. Expanding Δ = |x₁, y₁, 1; x₂, y₂, 1; x₃, y₃, 1| along column 3 gives exactly this bracket. So area = (1/2)|Δ|. The determinant can come out negative (it depends on the order of the vertices), so always take its absolute value. Write the answer in square units.
Rules to remember
- Write 1 in the third column of every row.
- Take the absolute value; an area is never negative.
- Δ = 0 means the three points are collinear (they form no triangle).
- When the area is given, use Δ = ±2 × area and solve both equations.
- For the line through two points, put a general point (x, y) in row 1 and set the determinant equal to 0.
Example 1: find the area
Find the area of the triangle with vertices (2, 1), (5, 3) and (−1, 4).
- Δ = |2, 1, 1; 5, 3, 1; −1, 4, 1|
- Expand along row 1: 2(3 × 1 − 1 × 4) − 1(5 × 1 − 1 × (−1)) + 1(5 × 4 − 3 × (−1))
- = 2(3 − 4) − 1(5 + 1) + 1(20 + 3)
- = −2 − 6 + 23 = 15
- Area = (1/2) × |15| = 15/2
Area = 15/2 = 7.5 square units
Example 2: find k from the area
Find k if the area of the triangle with vertices (k, 0), (3, 0) and (0, 4) is 10 square units.
- Δ = |k, 0, 1; 3, 0, 1; 0, 4, 1|
- Column 2 is 0, 0, 4, so expand along column 2. The sign at (3, 2) is −.
- Δ = −4 × |k, 1; 3, 1| = −4(k − 3) = 12 − 4k
- Area = 10, so Δ = ±20.
- 12 − 4k = 20 gives k = −2; 12 − 4k = −20 gives k = 8.
- Check k = 8: the base from x = 3 to x = 8 is 5, the height is 4, and (1/2)(5)(4) = 10.
k = 8 or k = −2
Example 3: equation of a line
Using determinants, find the equation of the line through (1, 3) and (4, 9).
- Let (x, y) be any point on the line. The three points are collinear, so Δ = 0.
- |x, y, 1; 1, 3, 1; 4, 9, 1| = 0
- x(3 − 9) − y(1 − 4) + 1(9 − 12) = 0
- −6x + 3y − 3 = 0
- Divide by −3: 2x − y + 1 = 0
- Check (4, 9): 8 − 9 + 1 = 0. Correct.
2x − y + 1 = 0, that is y = 2x + 1
Land surveyors split an irregular plot into triangles and find each area from the coordinates of its corners, exactly as this formula does. GPS field-mapping apps for farm plots work the same way.
2 marks: “Show that the points A(1, −1), B(2, 1) and C(4, 5) are collinear.” Δ = |1, −1, 1; 2, 1, 1; 4, 5, 1| = 1(1 − 5) − (−1)(2 − 4) + 1(10 − 4) = −4 − 2 + 6 = 0. The area is 0, so the points are collinear.
2 marks: “Find k if the area of the triangle with vertices (−3, 0), (0, 2) and (0, k) is 6 square units.” Expand along column 1: Δ = −3(2 − k) = 3k − 6. (1/2)|3k − 6| = 6, so 3k − 6 = ±12. k = 6 or k = −2.
3 marks: “Using determinants, find the equation of the line joining (−2, 3) and (4, 6). Also find k if (k, 5) lies on this line.” |x, y, 1; −2, 3, 1; 4, 6, 1| = 0 gives −3x + 6y − 24 = 0, that is x − 2y + 8 = 0. Put in (k, 5): k − 10 + 8 = 0, so k = 2.
Half the mod of the det, with 1s in the last column. Zero area means one straight line.