Question

Using integration, find the area of the region bounded by the triangle ABC whose vertices A, B, C are (−1, 1), (0, 5) and (3, 2) respectively.

Answer 1

$$\begin{gathered} \mathrm{Equation}\mathrm{of}\mathrm{line}\mathrm{AB}\mathrm{is} \\ y-1=\left(\frac{5-1}{0+1}\right)\left(x-\left(-1\right)\right) \\ \Rightarrow y=4x+5 \\ \mathrm{Area}\mathrm{under}\mathrm{the}\mathrm{line}\mathrm{AB}=\mathrm{area}\mathrm{ABDO} \\ ={\int }_{-1}^0\left(4x+5\right)dx \\ ={\left[4\frac{x^2}{2}+5x\right]}_{-1}^0 \\ =0-\left(2-5\right) \\ \Rightarrow \mathrm{Area}\mathrm{ABDO}=3\mathrm{sq}.\mathrm{units}...\left(1\right) \\ \mathrm{Equation}\mathrm{of}\mathrm{line}\mathrm{BC}\mathrm{is} \\ y-5=\left(\frac{2-5}{3-0}\right)\left(x-0\right) \\ \Rightarrow y=-x+5 \\ \mathrm{Area}\mathrm{under}\mathrm{line}\mathrm{BC}=\mathrm{Area}\mathrm{OBCP} \\ ={\int }_0^3\left(-x+5\right)dx \\ ={\left[-\frac{x^2}{2}+5x\right]}_0^3 \\ =-\frac{9}{2}+15-0 \\ \Rightarrow \mathrm{Area}\mathrm{OBCP}=\frac{21}{2}\mathrm{sq}.\mathrm{units}...\left(2\right) \\ \mathrm{Equation}\mathrm{of}\mathrm{line}\mathrm{CA}\mathrm{is} \\ y-2=\left(\frac{2-1}{3-\left(-1\right)}\right)\left(x-3\right) \\ \Rightarrow 4y=x+5 \\ \therefore \mathrm{Area}\mathrm{under}\mathrm{line}\mathrm{AC}=\mathrm{Area}\mathrm{ACPA} \\ \Rightarrow A={\int }_{-1}^3\left(\frac{x+5}{4}\right)dx \\ \Rightarrow A=\frac{1}{4}{\left[{\frac{x}{2}}^2+5x\right]}_{-1}^3 \\ \Rightarrow A=\frac{1}{4}\left[{\frac{3}{2}}^2+5\times 3-{\frac{\left(-1\right)}{2}}^2+5\left(-1\right)\right] \\ \Rightarrow A=\frac{1}{4}\left[\frac{9}{2}+15-\frac{1}{2}+5\right] \\ \Rightarrow \mathrm{Area}\mathrm{ACPA}=\frac{24}{4}=6\mathrm{sq}.\mathrm{units}...\left(3\right) \\ \mathrm{From}\left(1\right),\left(2\right)\mathrm{and}\left(3\right) \\ \mathrm{Area}\mathrm{\Delta }\mathrm{ABC}=\mathrm{Area}\mathrm{ABDO}+\mathrm{Area}\mathrm{OBCP}-\mathrm{Area}\mathrm{ACPA} \\ \Rightarrow A=3+\frac{21}{2}-6 \\ \Rightarrow A=\frac{21}{2}-3=\frac{21-6}{2}=\frac{15}{2}\mathrm{sq}.\mathrm{units} \\ \therefore \mathrm{Area}\mathrm{\Delta }\mathrm{ABC}=\frac{15}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$