IIT JEE Main Maths -Unit 3- Area of triangle using determinants- Study Notes-New Syllabus

Area of a Triangle Using Determinants

The area of a triangle whose vertices are \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) can be calculated using determinants.

$ \text{Area} = \dfrac{1}{2} \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix} $

Since area is always positive, we take the absolute value:

$ \text{Area} = \dfrac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix} \right| $

 Expansion Formula

Expanding the determinant along the first row:

$ \text{Area} = \dfrac{1}{2} |x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)| $

This is a commonly used formula in coordinate geometry to find the area directly from coordinates.

Collinearity Condition

If the three points are collinear, the area of the triangle formed by them is zero.

$ \begin{vmatrix} x_1 & y_1 & 1\\ x_2 & y_2 & 1\\ x_3 & y_3 & 1 \end{vmatrix} = 0 $

This determinant condition is often used in JEE questions to check if three points lie on the same straight line.

Geometrical Meaning

• The determinant gives twice the signed area of the triangle.
• The sign of the determinant indicates orientation (clockwise or anticlockwise order of vertices).
• Area is always taken as positive by using modulus.