[Maths Class Notes] on Determinant to Find the Area of a Triangle Pdf for Exam

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 ).

 

()

 

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}  & 1end {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}  & 1end {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  & 1end {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}  & 1end {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  & 1end {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}  & 1end {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  & 1end {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