 # Area of a Triangle Using Determinants

Imagine a triangle with vertices at (x1,y1), (x2,y2), and (x3,y3). If the triangle was a right-angled triangle, it would be pretty easy to compute the area of a triangle by finding one-half the product of the base and the height (area of triangle formula). However, when the triangle is not a right-angled triangle there are multiple different ways to do so. It turns out that the area of triangle formula can also be found using determinants. Let us see in detail how do we go about it.

### Suggested Videos        Properties of Determinants Applications of Determinants Minors and Cofactors_H ## Area of Triangle Formula Using Determinants

In earlier classes, we have studied that the area of a triangle whose vertices are  (x1, y1), (x2, y2) and (x3, y3), is given by the expression $$\frac{1}{2} [x1(y2–y3) + x2 (y3–y1) + x3 (y1–y2)]$$. Now this expression can be written in the form of a determinant as ### Points to be Noted

• Since the area is a positive quantity, we always take the absolute value of the determinant in (1).
• If the area is given, uses both positive and negative values of the determinant for calculation.
• The area of the triangle formed by three collinear points is zero.

Example: Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1).
Solution: The area of the triangle is given by = $$\frac{1}{2} [ 3(2 –1) – 8(– 4 – 5) + 1(– 4 –10)]$$
= $$\frac{1}{2} ( 3 + 72 – 14 ) = 61/2$$

## Derivation of the Area of Triangle Formula • We know that the Area of Rectangle can be written as follows • Area of Triangle A: Technically, each of those distances should be the absolute value of the difference. But the problem is much easier to work without the absolute values. • Area of Triangle B: Realize, however, as if the points don’t lie in the same positions (point 2 is both the rightmost and the uppermost), that the area found using these formulas will be negative. • Area of Triangle C: For that reason, caution should be exerted to always make the final answer non-negative. The area of a triangle, after all, can’t be negative. ### Let’s add the areas of the three outside triangles together. Simplifying further, Now, to subtract the areas of the three triangles from the area of the rectangle. Simplifying further, Let’s regroup those terms ………….. (1)

Now, consider the determinant formed by placing the x-coordinates in the first column, the y-coordinates in the second column, and the constant 1 in the last column. Let’s evaluate the determinant by expanding along the 3rd column. …………… (2)

On comparing both (1) and (2) we notice that the area of the triangle differs only in sign. The reason for this is because of the order the points were chosen in. If the points were chosen to be points 1, 2, and 3 in a different order, then the determinant would change only in sign.

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## Solved Examples on Area of Triangle Formula

Question: If the lines p1x + q1y = 1, p2x + q2y = 1 and p3x + q3y = 1 be concurrent, then the points (p1,q1), (p2,q2) and (p3,q3) ,

1. form scalene triangle
2. form equilateral triangle
3. are collinear
4. form a right-angled triangle

Solution: p1q11, p2q21 p3q31. Given lines are concurrent $$\begin{vmatrix} p_{1} & q_{1} & 1\\ p_{2} & q_{2} & 1 \\ p_{3} & q_{3} & 1\end{vmatrix} = 0$$

The left-hand side of the above equation is also equal to twice the area of a triangle with coordinates (p1,q1)(p2,q2)(p3,q3). From the area of triangle formula, and since the area is equal to zero, (p1,q1)(p2,q2)(p3,q3are collinear.

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jannat

Is determinant available just for square matrix?

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