Area of A Triangle using Determinants | Applied Mathematics for Class 12 - Commerce PDF Download

Determinant To Find Area of A Triangle

We know what a determinant is, let us know how to use Determinant to find Area of a Triangle
Suppose we are given three points in the Cartesian plane as (x₁, y₁), (x₂, y₂) and (x₃, y₃). The area of the triangle obtained by joining these points is given by,

Where α denotes the area of the triangle and (x₁, y₁), (x₂, y₂) and (x₃, y₃), represent the vertices of the triangle.

The formula for finding area could be represented in the form of determinants as given below.

The determinant represents the signed area of a triangle formed by three points. To ensure a positive area, we take the absolute value of the determinant. If the area is known, both positive and negative determinants are considered. If the points are collinear, forming a straight line, the determinant is zero, indicating zero area. To calculate the area using determinant expansion techniques, minors and cofactors are employed.

Therefore,

Hence, we see how determinants are applied to make calculations easy. Now let us try our hands at this application of determinants to find out the area of triangles.

Example1: Find out the area of the triangle whose vertices are given by A(0,0) , B (3,1) and C (2,4).
Sol: Using determinants we can easily find out the area of the triangle obtained by joining these points using the formula

Substituting the respective values in the determinant we have

Expanding the above determinant by using expansion techniques of determinant we get,
α = 1/2 [0(1−4)–0(3−2)+1(12−2)]
⇒ α = 5units

Example2: Find the area of the triangle whose vertices are A ( 1, 1 ), B ( 4, 2 ), and C ( 3, 5)
Sol: Using the formula that we have previously learnt, we can  find out the area of the triangle by joining the point given in the formula

When you substitute the given values in the above formula, we get:
k = ½ (1 ( 2 - 5 ) - 4 ( 4 - 3 ) + 3 ( 20 - 3 ))
k = ½ (1 ( -3 ) -4 ( 1 ) + 3 ( 17 ))
k = ½ (- 3 - 4 + 51)
k = ½ (44)
k = 22 units.
Since the area of the triangle cannot be negative, the value of k = 3 units.