Question

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are (-6, 4), (-4, -10) and (18, -2).

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{A}\equiv \left(-6,4\right) \\ \mathrm{B}\equiv \left(-4,-10\right) \\ \mathrm{and}\mathrm{C}\equiv \left(18,-2\right) \\ \mathrm{P},\mathrm{Q}\mathrm{and}\mathrm{R}\mathrm{are}\mathrm{the}\mathrm{mid}\mathrm{points}\mathrm{of}\mathrm{AB},\mathrm{BC}\mathrm{and}\mathrm{AC},\mathrm{respectively}. \\ \mathrm{Let}\mathrm{P}\equiv \left(x_1,y_1\right)\therefore x_1=\frac{-6-4}{2}=-5,y_1=\frac{4-10}{2}=-3 \\ \mathrm{Q}\equiv \left(x_2,y_2\right)\therefore x_2=\frac{-4+18}{2}=7,y_2=\frac{-10-2}{2}=-6 \\ \mathrm{and}\mathrm{R}\equiv \left(x_3,y_3\right)\therefore x_3=\frac{18-6}{2}=6,y_3=\frac{4-2}{2}=1 \end{gathered}$$

$$\begin{gathered} \mathrm{Area}\mathrm{of}\text{a}\mathrm{triangle}=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ \mathrm{Area}\mathrm{of}△\mathrm{PQR}=\frac{1}{2}\left[-5\left(-6-1\right)+7\left\{1-\left(-3\right)\right\}+6\left\{-3-\left(-6\right)\right\}\right] \\ =\frac{1}{2}\left[-5\left(-7\right)+7\left(4\right)+6\left(3\right)\right] \\ =\frac{1}{2}\left(35+28+18\right) \\ =\frac{1}{2}\times 81=40.5\text{square units} \end{gathered}$$