Determinant to Find the Area of a Triangle

Area of Triangle Using Determinant

We have previously learnt that 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 value or the absolute value of the determinant that is obtained. 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 if 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. 


Example: 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.


Solved Examples  


Question 1: 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


Question 2: 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


FAQ (Frequently Asked Questions)

1. How do you calculate the area of a triangle with the determinant?

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 ( y- y1 ) + x3 ( y1 - y2 ) ]


Here, k is the area of the triangle and the vertices of the triangle are represented by: (x1, y1), (x2, y2 ), and ( x3, y3 ).In order to find the are of the triangle with the help of a determinant, you use the formula given below:


 Image will be uploaded soon


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 value or the absolute value of the determinant that is obtained. In case we already know the area of the triangle or the area has been given in the equation