Coordinate Geometry • Topic 4 of 4

Area of Triangle

Given the three vertices A(x₁, y₁), B(x₂, y₂), C(x₃, y₃), the area of the triangle is ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|. The absolute value is essential — the bracketed expression can come out negative (it is really a signed determinant) and area is never negative, so always take the magnitude. The single most useful corollary for CAT: if this expression equals zero, the three points are collinear, because a "triangle" of zero area is really three points on one line. That makes the area formula the quickest collinearity test there is — much faster than computing two slopes and comparing. A clean way to set up the computation by hand is the shoelace pattern: write the three points in a column, repeat the first at the bottom, cross-multiply down-right minus down-left, sum, halve and take the modulus. Two practical CAT moves: when one vertex is the origin the formula shrinks to ½|x₁y₂ − x₂y₁|, and when a question gives the area and leaves one coordinate unknown, set the determinant equal to ±2·(area) to get two possible values for that coordinate.

✅ Solved examples

1. Find the area of the triangle with vertices (0, 0), (4, 0) and (0, 3).
½|0(0−3) + 4(3−0) + 0(0−0)| = ½|12| = 6. (A right triangle with legs 4 and 3.)
2. Find the area of the triangle with vertices (1, 1), (4, 2) and (3, 5).
½|1(2−5) + 4(5−1) + 3(1−2)| = ½|−3 + 16 − 3| = ½·10 = 5.
3. Show that the points (1, 4), (3, 8) and (5, 12) are collinear.
½|1(8−12) + 3(12−4) + 5(4−8)| = ½|−4 + 24 − 20| = ½·0 = 0. Area zero ⇒ the three points are collinear.
4. The area of the triangle with vertices (1, 2), (3, k) and (5, 6) is 4. Find k.
½|1(k−6) + 3(6−2) + 5(2−k)| = 4 ⇒ |k − 6 + 12 + 10 − 5k| = 8 ⇒ |16 − 4k| = 8 ⇒ 16 − 4k = ±8 ⇒ k = 2 or k = 6.

✏️ Practice — try these, take hints as needed

1. Find the area of the triangle with vertices (0, 0), (5, 0) and (2, 4).
Origin vertex ⇒ use ½|x₁y₂ − x₂y₁| on the other two.
Or just base × height.
Base 5, height 4.
10
2. Find the area of the triangle with vertices (2, 3), (−1, 0) and (2, −4).
Apply the determinant formula.
½|2(0−(−4)) + (−1)((−4)−3) + 2(3−0)|.
½|8 + 7 + 6|.
10.5
3. For what value of p are (2, 1), (p, 4) and (4, 7) collinear?
Set the area determinant to 0.
2(4−7) + p(7−1) + 4(1−4) = 0.
−6 + 6p − 12 = 0.
p = 3
4. The triangle with vertices (k, 0), (0, 4) and (2, 0) has area 10. Find k.
Base on the x-axis = |k − 2|, height = 4.
½·|k−2|·4 = 10.
|k − 2| = 5.
k = 7 or k = −3
5. Find the area of the quadrilateral with vertices (1, 1), (4, 1), (4, 5) and (1, 5).
It is a rectangle.
Width 3, height 4.
Area = width × height.
12

📝 Topic test — 8 questions

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