Area of Triangle Using Determinant
Before understanding the determinant for finding the area of a triangle, let us have a quick look over the meaning of determinant first. So, the sum-product which is obtained by the elements of the square matrix is called a determinant. It helps to find the adjoint of the matrix, as well as the inverse of the matrix.
Now, finding the area of a triangle is not that difficult if the given triangle is a right-angle triangle, because the area of such a triangle can easily be found by finding the product, one-half, of the base and the height. But if the triangle is not the right-angle triangle, then finding the area of the triangle is not that easy.
Hence, there are few other methods of finding the area in such cases and one such method is finding the area of a triangle using the determinants.
The determinant is the scalar value which is computed from different elements of a square matrix that has certain properties of a linear transformation. Let us now learn how to use the determinant to find the area of a triangle. Let’s say that (x1, y1), (x2, y2 ), and ( x3, y3 ) are three points of the triangle in the cartesian plane.
Now the area of the triangle of the will be given as:
k = ½ [ x1 ( y2 - y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 ) ]
Here, k is the area of the triangle using determinant and the vertices of the triangle are represented by (x1, y1), (x2, y2 ), and ( x3, y3 ).
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In order to find the area of a triangle in determinant form, you use the formula given below:
K = ½ \[\begin{bmatrix} x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1 \\ x_{3} & y_{1} & 1\end {bmatrix}\]
The value of the determinant is either positive or negative but since here we are talking about the area of the triangle, we cannot have a negative value. Hence, we take the positive and the negative value or the absolute value of the determinant.
In case we already know the area of the triangle or the area has been given in the equation, we can use both the positive values of the determinant and the negative value of the determinant. In case three points are colinear, then it forms a line and not a triangle and the area of the triangle that is enclosed in a straight line is equal to 0. Therefore, the value of the determinant to find the area of the triangle would also be equal to zero. Keeping the aforementioned statements in mind, let us use the determinant expansion techniques using minors and cofactors and try to expand the determinant which denotes the area of the triangle.
Hence,
k = ½ (x1 ( y2 - y3 ) + x2 ( y3 - y1 ) + x3 ( y1 - y2 ))
This is how you apply determinants to make the calculation of the determinant easy. Let us apply this to a matrix and understand the concept much better.
Solved Examples
1. Find the area of the triangle whose vertices are A ( 1, 1 ), B ( 4, 2 ), and C ( 3, 5)
Solution: Using the formula that we have previously learnt, we can find out the area of the triangle by joining the point given in the formula
K = ½ \[\begin{bmatrix} x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1 \\ x_{3} & y_{1} & 1\end {bmatrix}\]
When you substitute the given values in the above formula, we get:
K = ½ \[\begin{bmatrix} 1 & 1 & 1\\ 4 & 2 & 1 \\ 3 & 5 & 1\end {bmatrix}\]
k = ½ (1 ( 2 - 5 ) - 4 ( 4 - 3 ) + 3 ( 20 - 3 ))
k = ½ (1 ( -3 ) -4 ( 1 ) + 3 ( 17 ))
k = ½ (- 3 - 4 + 51)
k = ½ (44)
k = 22 units.
Since the area of the triangle cannot be negative, the value of k = 3 units.
2. Find the area of a triangle by determinant method whose vertices are A ( 4, 9 ), B ( - 3, 3 ), and C ( 6, 2 )
Solution: Using the formula that we have previously learnt, we can find out the area of the triangle by joining the point given in the formula
K = ½ \[\begin{bmatrix} x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1 \\ x_{3} & y_{1} & 1\end {bmatrix}\]
When you substitute the given values in the above formula, we get:
K = ½ \[\begin{bmatrix} 4 & 9 & 1\\ -3 & 3 & 1 \\ 6 & 2 & 1\end {bmatrix}\]
k = ½ (4 ( 3 - 2) - 9 ( -3 - 6 ) + 1 ( - 6 - 18 ))
k = ½ (4 ( 1 ) - 9 ( - 9 ) +1 ( - 24 ))
k = ½ (4 + 81 - 24)
k = ½ (61)
k = 61 / 2 units
3. Find the area of the triangle whose vertices are A ( 4, 8 ), B ( - 6, 2 ), and C ( 5, 7 )
Solution: Using the formula that we have previously learnt, we can find out the area of the triangle by joining the point given in the formula
K = ½ \[\begin{bmatrix} x_{1} & y_{1} & 1\\ x_{2} & y_{2} & 1 \\ x_{3} & y_{1} & 1\end {bmatrix}\]
When you substitute the given values in the above formula, we get:
K = ½ \[\begin{bmatrix} 4 & 8 & 1\\ -6 & 2 & 1 \\ 5 & 7 & 1\end {bmatrix}\]
k = ½ (4 ( 2 - 7 ) - 8 ( - 6 - 5 ) + 1 ( - 42 - 10 ))
k = ½ (4 ( - 5 ) - 8 ( - 11 ) +1 ( - 52 ))
k = ½ (20 + 88 - 52)
k = ½ (56)
k = 28 units
FAQs on Determinant to Find the Area of a Triangle
1. What is the step-by-step method to solve for the area of a triangle with given vertices using determinants, as per the NCERT Class 12 syllabus?
To find the area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) using the determinant method, follow these steps as per the CBSE 2025-26 guidelines:
- Step 1: Set up a 3x3 matrix where the first column contains the x-coordinates, the second column contains the y-coordinates, and the third column is filled with 1s.
- Step 2: Place this matrix inside the area formula: Area = ½ |det(A)|, where A is the matrix.
- Step 3: The formula is written as: Area = ½ | x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂) |.
- Step 4: Calculate the value of the determinant.
- Step 5: Take the absolute value of the result and multiply by ½ to get the final area in square units. Since area cannot be negative, we only consider the positive magnitude.
2. How can you use the determinant method to check if three points are collinear?
The correct method to check for collinearity using determinants is based on the concept of area. Three points are collinear if they lie on the same straight line. A triangle formed by three collinear points would have an area of zero. Therefore, to check if points (x₁, y₁), (x₂, y₂), and (x₃, y₃) are collinear, you calculate the area using the determinant formula. If the value of the determinant Δ = ½ [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] is zero, the points are collinear.
3. Why is the absolute value of the determinant taken when calculating the area of a triangle?
The value of a determinant can be positive or negative depending on the order of the vertices (clockwise or counter-clockwise). However, the area of a triangle is a geometric measure of space and must always be a non-negative quantity. Taking the absolute value, or modulus, of the determinant's result ensures that the final calculated area is always positive, which aligns with the physical meaning of area.
4. How is the formula for the area of a triangle using determinants derived from the coordinate geometry formula?
The determinant formula is essentially a compact and systematic way of writing the standard coordinate geometry (or shoelace) formula. The standard formula is: Area = ½ |(x₁ (y₂ - y₃) + x₂ (y₃ - y₁) + x₃ (y₁ - y₂))|. When you set up the 3x3 matrix for the area of a triangle and calculate its determinant, the expansion along the first column (or any row/column) yields this exact expression. The determinant provides a procedural structure to arrive at the same calculation.
5. What is a common mistake when solving for an unknown coordinate if the area of the triangle is already given?
A common pitfall is forgetting that the determinant can have both a positive and a negative value. If the area of a triangle is given as 'k' square units, you must set the determinant expression (without the absolute value) equal to both +k and -k. For example, if Area = 4 sq. units, you must solve for the unknown by equating the expression ½ [x₁(y₂ – y₃) + ...] to both +4 and -4. Solving for only the positive case will result in finding only one of the possible solutions for the unknown coordinate.
6. How do you find the equation of a line passing through two given points using the determinant method?
To find the equation of a line passing through two points, A(x₁, y₁) and B(x₂, y₂), you can use the concept of collinearity. Let P(x, y) be any point on the line passing through A and B. Since all three points are collinear, the area of the triangle formed by A, B, and P must be zero. You set up the determinant for the area and equate it to zero:
½ | x(y₁ - y₂) + x₁(y₂ - y) + x₂(y - y₁) | = 0.
Simplifying this equation gives you the required linear equation of the line.
