Area of a Triangle - Determinants

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.

The area of a triangle is half the product of the base and altitude of the triangle. But if the height of the triangle is unknown and its vertices are given, then we can find the area of the triangle using the determinant formula.

In this article, we will calculate the area of a triangle in determinant form using its formula.

Area of a Triangle

Triangle - Determinant

The area of triangle in determinant form can be calculated if the vertices of the triangle are given. Consider the triangle $ABC$ with vertices $A(x_{1},y_{1})$, $B(x_{2},y_{2})$ , $C(x_{3},y_{3})$ then its area can be calculated as

${\frac {1}{2}}\left[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right]$

This expression in the determinant form is written as

$\bigtriangleup ={\frac {1}{2}}{\begin{vmatrix}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{vmatrix}}..............(1)$

$\bigtriangleup ={\frac {1}{2}}\left[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\right]$

• Since area is a positive quantity, we always take the absolute value of the determinant in (1).

• If area is given, use both positive and negative values of the determinant for calculation.
• The area of the triangle formed by three collinear points is zero

Example

Find the area of a triangle whose vertices are $(3,8)$,$(-4,2)$ ,$(5,1)$

The Area of of triangle is

$\bigtriangleup ={\frac {1}{2}}{\begin{vmatrix}3&8&1\\-4&2&1\\5&1&1\end{vmatrix}}$

$\bigtriangleup ={\frac {1}{2}}\left[3(2-1)-8(-4-5)+1(-4-10)\right]$

$\bigtriangleup ={\frac {1}{2}}\left[3(1)-8(-9)+1(-14)\right]$

$\bigtriangleup ={\frac {1}{2}}\left[3+72-14\right]$

$\bigtriangleup ={\frac {1}{2}}\left[61\right]$

$\bigtriangleup ={\frac {61}{2}}$