4.3 Area of a Triangle

In earlier classes, we have studied that the area of a triangle whose vertices are \((x_{1},y_{1})\), \((x_2, y_2)\) and \((x_{3},y_{3})\), is given by the expression \(\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+ x_{3}(y_{1}-y_{2})]\) Now this expression can be written in the form of a determinant as \(A=\frac{1}{2}|\begin{smallmatrix}x_{1}&y_{1}&1\\ x_{2}&y_{2}&1\\ x_{3}&y_{3}&1\end{smallmatrix}|\)

Remarks

  1. Since area is a positive quantity, we always take the absolute value of the determinant in (1).
  2. If area is given, use both positive and negative values of the determinant for calculation.
  3. The area of the triangle formed by three collinear points is zero.