Area of Triangle in Determinant Form – Definition, Examples

What Is the Area of a Triangle in Determinant Form?

In coordinate geometry, the area of a triangle in determinant form can be calculated when the coordinates of the vertices of the triangle are provided. Calculating the area of a triangle in determinant form is an important application of determinants.

If the vertices of a triangle PQR are P(x₁, y₁), Q(x₂, y₂), and R(x₃, y₃), the area of the triangle can be calculated as follows:

Note that the process of expanding to find the determinant of a matrix remains consistent regardless of the chosen row or column. The result remains the same, making the choice of row or column arbitrary.

Expanding the determinant along the first column, we get

Area $= \frac{1}{2} \times \left[x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})\right]$

A determinant of a square matrix A is a real value denoted by | A | or det (A). It is a scalar value held by every square matrix. 

Formula of Area of Triangle in Determinant Form

Normally, the area of a triangle is calculated using the formula $\frac{1}{2} \times b \times h, where b is the base and h is height of the triangle. 

Some cases, however, use a different formula for determining the triangle’s area where the height is not given. The vertices of the triangle are employed for the evaluation of the area using the following determinant formula.

Suppose a triangle ΔPQR has vertices P(x₁, y₁), Q(x₂, y₂), and R(x₃, y₃). 

The area of ΔPQR is given by

Solving the determinant, we get

How to Determine the Area of Triangle in Determinant Form

Now that we know the formula to evaluate the area of triangle in determinant form, we will solve a few examples to understand its application. 

A determinant value can be both positive or negative. However, area can never be negative. So, we take the absolute value of the result.

Example: Suppose the vertices are P(-1, 2), Q(3, 1), and R(2, 5). So, if we want to determine the area of triangle in determinant form, we will use the above-mentioned formula.

Area $= \frac{1}{2} \times \left[x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})\right]$

Here, we know that the coordinates of vertices are P(x₁, y₁), Q(x₂, y₂), and R(x₃, y₃).

x₁= -1, y₁= 2

x₂= 3, y₂= 1

x₃= 2, y₃= 5

Putting the values in the formula, we get

A $= (\frac{1}{2}) \left[(-1) (1 – 5) + 3 (5 – 2) + 2 (2 – 1)\right]$

A $= \frac{1}{2} \times 15$

A $= \frac{15}{2}$ square units

Facts about Area of Triangle in Determinant Form

Conclusion

In this article, we delved into the concept of finding the area of a triangle using the determinant form, a powerful method in linear algebra. We learned how determinants are used to find the area of a triangle, the formula and examples. To reinforce our knowledge, let’s engage in solving examples and attempting MCQs for better comprehension. 

Solved Examples on Area of Triangle in Determinant Form

1. The three vertices of ▵GHI are G(–1, 1), H(3, 4), and I(2, –1). How can we find the area of this triangle?

Solution:

In a case where we just have the vertices given, we can find the area by using the determinant formula.

x₁ = –1, y₁ = 1

x₂ = 3, y₂ = 4

x₃ = 2, y₃ = -1

So, let us use the determinant formula.

Area $= \frac{1}{2} \times \left[x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})\right]$

Putting the values of the vertices in the formula, we get

A $= \frac{1}{2} \times \left[(-1) (4 – (-1)) + 3 (-1 – 1) + 2 (1 – 4)\right]$

A $= \frac{1}{2} \times \left[-5 – 6 – 6\right]$

A $=  \frac{-17}{2}$ sq units

Therefore, the area of the triangle GHI is 3.5 sq units.

2. Use a determinant to find the area of the triangle with the given vertices: A(2, 4), B(5, 6), and C(1, 3). 

Solution:

x₁ = 2, y₁ = 4

x₂ = 5, y₂ = 6

x₃ = 1, y₃ = 3

So, let us use the determinant formula.

Area $= \frac{1}{2} \times \left[x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})\right]$

Putting the values in the formula, we get

Area $= \frac{1}{2} \times \left[2 (6 – 3) + 5 (3 – 4) + 1 (4 -6)\right]$

Area $= \frac{1}{2} \times |  -1 | $

Area = 0.5 square unit

Therefore, the area of the triangle ABC is 0.5 square unit.

3. The vertices of a triangle are given as P(0, 0), Q(2, 4), and R(5, 1). What would be the area of this triangle?

Solution:

x₁= 0, y₁= 0

x₂= 2, y₂= 4

x₃= 5, y₃= 1

A $= \frac{1}{2} \times \left[x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})\right]$

Putting the values of the vertices in the formula, we get

A $= (\frac{1}{2}) \times \left[0 (4 – 1) + 2 (1 – 0) + 5 (0 – 4)\right]$

A $= (\frac{1}{2}) \times |-18|$

A = 9

Practice Problems of Area of Triangle in Determinant Form

Frequently Asked Questions about the Area of Triangle in Determinant Form