Question

Find the area of the triangle whose vertices are given below:
(i) (-1, 5), (-2, -3), (10, 1)

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{A}\equiv \left(-1,5\right)\therefore x_1=-1,y_1=5 \\ \mathrm{B}\equiv \left(-2,-3\right)\therefore x_2=-2,y_2=-3 \\ \mathrm{and}\mathrm{C}\equiv \left(10,1\right)\therefore x_3=10,y_3=1 \end{gathered}$$

$$\begin{gathered} \mathrm{Area}\mathrm{of}\text{a}\mathrm{triangle}=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ \mathrm{Area}\mathrm{of}△\mathrm{ABC}=\frac{1}{2}\left[-1\left(-3-1\right)-2\left(1-5\right)+10\left\{5-\left(-3\right)\right\}\right] \\ =\frac{1}{2}\left[-1\left(-4\right)-2\left(-4\right)+10\left(8\right)\right] \\ =\frac{1}{2}\left(4+8+80\right) \\ =\frac{1}{2}\times 92=46\mathrm{square}\mathrm{units} \end{gathered}$$