Area of Triangle in Determinant Form
The area of triangle in determinant form is calculated in coordinate geometry when the coordinates of the vertices of the triangle are given. Finding the area of triangle in determinant form is one of the important applications of determinants. Generally, we determine the area of a triangle using the formula half the product of the base and altitude of the triangle. But if the height of the triangle is not known and its vertices are given, then we can find the area of the triangle using the determinant formula.
In this article, we will compute the area of a triangle in determinant form using its formula. We will solve a few examples based on the concept of the area of triangle in determinant form for a better understanding of its application.
What is the Area of Triangle in Determinant Form?
The area of triangle in determinant form can be evaluated if the vertices of the triangle are given. If we have a triangle ABC with vertices A(x_{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}), then its area can be calculated as (1/2) [x_{1 }(y_{2}  y_{3}) + x_{2 }(y_{3}  y_{1}) + x_{3 }(y_{1}  y_{2})]. The general formula for the area of a triangle is half the product of its base and height. But in coordinate geometry, if the coordinates of the vertices of a triangle are given, then we can compute the area of triangle in determinant form. Let us now go through its formula in the determinant form in the next section.
Formula of Area of Triangle in Determinant Form
The formula for the area of a triangle in determinant form gives a scalar value that can be positive or negative. But since the area of a triangle can never be negative, we consider the absolute value of the determinant as the area of the triangle. Now, if the vertices of a triangle ABC are A(x_{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}) in a cartesian plane, then the area of triangle in determinant form formula is given by,
Area of triangle ABC = \(\dfrac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\)
After simplification, this formula of area of triangle in determinant form can also be written as:
Area of triangle ABC = (1/2)  [x_{1 }(y_{2}  y_{3}) + x_{2 }(y_{3}  y_{1}) + x_{3 }(y_{1}  y_{2})] 
How to Determine Area of Triangle in Determinant Form?
Now that we know the formula to evaluate the area of triangle in determinant form, we will now solve a few examples to understand its application. Let us consider a triangle ABC whose vertices with coordinates are given.
Example 1: Find the area of triangle in determinant form whose vertices are A(0, 0), B(0, 5), and C(8, 0).
Solution: To find the area of triangle ABC, we will use the formula, Area of triangle ABC = \(\dfrac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\)
Here x_{1} = 0, x_{2} = 0, x_{3} = 8, y_{1} = 0, y_{2} = 5, y_{3} = 0. Therefore, the area is given by,
Area = \(\dfrac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\)
= \(\dfrac{1}{2}\begin{vmatrix}0&0&1\\0&5&1\\8&0&1\end{vmatrix}\)
= (1/2)  [0_{ }(5  0) + 0_{ }(0  0) + 8_{ }(0  (5))] 
= (1/2)  0 + 0 + 8 × 5 
= (1/2) × 40
= 20 square units
Answer: Hence, the area of triangle ABC with vertices A(0, 0), B(0, 5), and C(8, 0) is 20 square units.
Example 2: Compute the area of a triangle PQR by the determinant method whose vertices are P(3, 5), Q(8, 5), and R(1, 2).
Solution: To find the area of triangle PQR, we will use the formula, Area of triangle PQR = \(\dfrac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\)
Here x_{1} = 3, x_{2} = 8, x_{3} = 1, y_{1} = 5, y_{2} = 5, y_{3} = 2. Therefore, the area is given by,
Area = (1/2)  [x_{1 }(y_{2}  y_{3}) + x_{2 }(y_{3}  y_{1}) + x_{3 }(y_{1}  y_{2})] 
= (1/2)  [3_{ }(5  2) + (8)_{ }(2  (5)) + 1_{ }(5  5)] 
= (1/2)  [3_{ }(3)  8_{ }(7) + 1_{ }(10)] 
= (1/2)  9  56  10
= (1/2) 75
= (1/2) × 75
= 37.5 square units
Answer: Hence, the area of triangle PQR is equal to 37.5 square units.
Important Notes on Area of Triangle in Determinant Form
 On simplification, the formula for the area of triangle ABC with vertices A(x_{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}) in determinant form is (1/2)  [x_{1 }(y_{2}  y_{3}) + x_{2 }(y_{3}  y_{1}) + x_{3 }(y_{1}  y_{2})] .
 Evaluating the area of triangle in determinant form is an application of determinants.
☛ Related Topics:
Area of Triangle in Determinant Form Examples

Example 1: Find the area of triangle in determinant form whose vertices are A(9, 3), B(3, 2), and C(7, 6).
Solution: To find the area of triangle ABC, we will use the formula, Area of triangle ABC = \(\dfrac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\)
Here x_{1} = 9, x_{2} = 3, x_{3} = 7, y_{1} = 3, y_{2} = 2, y_{3} = 6. Therefore, the area is given by,
Area = \(\dfrac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\)
= \(\dfrac{1}{2}\begin{vmatrix}9&3&1\\3&2&1\\7&6&1\end{vmatrix}\)
= (1/2)  [9_{ }(2  6) + 3_{ }(6  3) + (7)_{ }(3  2)] 
= (1/2)  9 × (4) + 3 × 3  7 × 1 
= (1/2) × 36 + 9  7
= (1/2) × 34
= 17 square units
Answer: Hence, the area of triangle ABC with vertices A(9, 3), B(3, 2), and C(7, 6) is 17 square units.

Example 2: Compute the area of a triangle PQR by the determinant method whose vertices are P(1, 4), Q(3, 4), and R(4, 5).
Solution: To find the area of triangle PQR, we will use the formula, Area of triangle PQR = \(\dfrac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\)
Here x_{1} = 1, x_{2} = 3, x_{3} = 4, y_{1} = 4, y_{2} = 4, y_{3} = 5. Therefore, the area is given by,
Area = (1/2)  [1_{ }(4  5) + 3_{ }(5  (4)) + 4_{ }(4  (4))] 
= (1/2)  [1_{ }(9) + 3_{ }(9) + 4_{ }(0)] 
= (1/2)  [9 + 27 + 0] 
= (1/2) 36
= (1/2) × 36
= 18 square units
Answer: Hence, the area of triangle PQR is equal to 18 square units.
FAQs on Area of Triangle in Determinant Form
What is Area of Triangle in Determinant Form?
The area of triangle in determinant form can be evaluated if the vertices of the triangle are given. If we have a triangle ABC with vertices A(x_{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}), then its area can be calculated as (1/2) [x_{1 }(y_{2}  y_{3}) + x_{2 }(y_{3}  y_{1}) + x_{3 }(y_{1}  y_{2})].
How Do You Find the Area of a Triangle With 3 Coordinates?
We can find the area of triangle with 3 coordinates using the determinant formula. For a triangle ABC with vertices A(x_{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}), we can use the formula \(\dfrac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\).
What is the Formula for Area of Triangle in Determinant Form?
The formula to find the area of triangle in determinant form using the formula \(\dfrac{1}{2}\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}\) where A(x_{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}) are the vertices of the triangle ABC. On simplification of the determinant formula, we have the formula as (1/2) [x_{1 }(y_{2}  y_{3}) + x_{2 }(y_{3}  y_{1}) + x_{3 }(y_{1}  y_{2})].
How Do You Find the Area of Triangle in Determinant Form?
The area of a triangle in determinant form can be determined using the formula (1/2) [x_{1 }(y_{2}  y_{3}) + x_{2 }(y_{3}  y_{1}) + x_{3 }(y_{1}  y_{2})], where A(x_{1}, y_{1}), B(x_{2}, y_{2}), and C(x_{3}, y_{3}) are the vertices of the triangle ABC.
visual curriculum