Area of a Triangle Using Determinant Formula with Solved Examples
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 Using Determinants to Calculate the Area of a Triangle
1. What is the formula for finding the area of a triangle using determinants?
The area of a triangle using determinants is given by Area = (1/2)|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.
- This formula is derived from the determinant of a 3 × 3 matrix.
- It is used when the coordinates of the three vertices are known.
- The absolute value ensures the area is always positive.
2. How do you find the area of a triangle using a 3×3 determinant?
To find the area of a triangle using a 3×3 determinant, use Area = (1/2)|det| where the determinant is formed from the coordinates.
- Write the matrix: |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|
- Find its determinant value.
- Take half of the absolute value of the result.
3. Why do we use absolute value in the determinant area formula?
We use the absolute value because the determinant can be positive or negative depending on the order of vertices, but area is always positive.
- A positive determinant indicates anticlockwise order.
- A negative determinant indicates clockwise order.
- The magnitude gives the actual geometric area.
4. Can you give an example of finding the area of a triangle using determinants?
Yes, for vertices (0,0), (4,0), and (0,3), the area using determinants is 6 square units.
- Form the determinant: |0 0 1; 4 0 1; 0 3 1|
- Determinant value = 12
- Area = (1/2) × |12| = 6
5. How does the determinant method prove that three points are collinear?
Three points are collinear if the determinant value is 0.
- Form the 3×3 determinant using the coordinates.
- If the area = (1/2)|det| = 0, then the points lie on the same straight line.
- This is a standard test for collinearity in coordinate geometry.
6. What is the matrix form used to find the area of a triangle?
The matrix form used is |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|.
- This is a 3 × 3 determinant.
- The third column consists of 1s.
- The area equals half of the absolute value of this determinant.
7. Is the determinant method for area valid for any triangle?
Yes, the determinant method works for any triangle in the coordinate plane as long as the three vertices are known.
- It applies to scalene, isosceles, and right triangles.
- It does not depend on side lengths directly.
- It only requires coordinate values.
8. What happens to the area if the order of points changes?
If the order of points changes, the determinant sign may change but the area remains the same.
- Clockwise order gives a negative determinant.
- Anticlockwise order gives a positive determinant.
- Taking absolute value ensures the same final area.
9. How is the determinant area formula related to the base and height formula?
The determinant formula is another form of the standard formula Area = (1/2) × base × height expressed in coordinate form.
- The determinant automatically calculates effective base and perpendicular height.
- It avoids manually finding distances and slopes.
- Both methods give the same numerical area.
10. What are common mistakes when using determinants to find the area of a triangle?
Common mistakes include calculation errors in expansion and forgetting the (1/2) factor.
- Incorrect determinant expansion.
- Not using absolute value.
- Omitting the multiplication by 1/2.
