Area of a Triangle

When the coordinates of the three vertices of a triangle are known, we can calculate its area without drawing it. Coordinate geometry gives us a direct formula to find the area, even if the triangle lies in any quadrant or has negative coordinates.

This method is especially useful when the triangle is drawn on a graph or when exact coordinates are given in problems.

If the vertices of a triangle are:

\(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\)

The area is given by:

\(\text{Area} = \dfrac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2 ) \right|\)

This formula works for all triangles — acute, obtuse, right, or even when sides cross axes.

This formula comes from the idea of finding the area using a determinant. It calculates the signed area of a parallelogram formed by the points and then takes half of it.

The absolute value (| |) ensures that the answer is always positive, since area cannot be negative.

If the three points lie on the same straight line (i.e., they are collinear), the area of the triangle formed is zero.

So, this formula can also test whether three points lie on the same line:

If area = 0 → points are collinear

Let us solve a few examples to understand the application of the formula.

• Not following the coordinate order correctly.
• Forgetting to apply the absolute value (| |).
• Incorrect subtraction inside the formula.
• Mixing up coordinates of different points.
• Thinking area must always be an integer (it can be decimal).

Find the area of triangles with the following vertices:

1. (0, 0), (4, 0), (4, 3)
2. (2, -1), (5, 6), (-3, 4)
3. (1, 2), (2, 4), (3, 6) — collinearity check
4. (-4, -2), (1, 3), (6, -1)

• Area of triangle using coordinates is found using determinant-based formula.
• Formula: 1/2 × | x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2 ) |
• If area = 0, the points are collinear.
• Useful for geometry problems without needing to draw the figure.