Question

Using integration, find the area of the triangle ABC coordinates of whose vertices are A (4, 1), B (6, 6) and C (8, 4).

Answer 1

A(4, 1), B(6, 6) and C(8, 4) are three given points.
$$\begin{gathered} \mathrm{Equation}\mathrm{of}AB\mathrm{is}\mathrm{given}\mathrm{by} \\ y-1=\frac{6-1}{6-4}\left(x-4\right) \\ \Rightarrow y-1=\frac{5}{2}\left(x-4\right) \\ \Rightarrow y=\frac{5}{2}x-9.....\left(1\right) \\ \mathrm{Equation}\mathrm{of}BC\mathrm{is}\mathrm{given}\mathrm{by} \\ y-6=\frac{4-6}{8-6}\left(x-6\right) \\ \Rightarrow y-6=-\left(x-6\right) \\ \Rightarrow y=-x+12.....\left(2\right) \\ \mathrm{Equation}\mathrm{of}CA\mathrm{is}\mathrm{given}\mathrm{by} \\ \Rightarrow y-4=\frac{1-4}{4-8}\left(x-8\right) \\ \Rightarrow y-4=\frac{3}{4}\left(x-8\right) \\ \Rightarrow y=\frac{3}{4}x-2.....\left(3\right) \end{gathered}$$
Required area of shaded region ABC
$$\begin{gathered} \mathrm{Shaded}\mathrm{area}\left(\mathit{A}\mathit{B}\mathit{C}\mathit{}\right)=\mathrm{area}\left(\mathit{A}\mathit{B}\mathit{D}\right)\hspace{0.17em}+\mathrm{area}\left(\mathit{D}\mathit{B}\mathit{C}\right) \\ \mathrm{Consider}\mathrm{a}\mathrm{point}\mathrm{P}\left(x,y_2\right)\mathrm{on}AB\mathrm{and}Q\left(x,y_1\right)\mathrm{on}AD \\ \Rightarrow \mathrm{for}\mathrm{a}\mathrm{vertical}\mathrm{strip}\mathrm{of}\mathrm{length}=\left|y_2-y_1\right|\mathrm{and}\mathrm{width}=dx,\mathrm{area}=\left|y_2-y_1\right|dx, \\ \mathrm{The}\mathrm{approximating}\mathrm{rectangle}\mathrm{moves}\mathrm{from}x=4\mathrm{to}x=6 \\ \mathrm{Hence}\mathrm{area}\left(ABD\right)={\int }_4^6\left|y_2-y_1\right|dx \\ ={\int }_4^6\left[\left(\frac{5}{2}x-9\right)-\left(\frac{3}{4}x-2\right)\right]dx \\ ={\int }_4^6\left(\frac{7}{4}x-7\right)dx \\ ={\left[\frac{7}{4}\times \frac{x^2}{2}-7x\right]}_4^6 \\ =\frac{7}{8}\left(36-16\right)-7\left(6-4\right) \\ =\frac{7}{8}\times 20-14 \\ =\frac{35}{2}-14 \\ =\frac{35-28}{2}=\frac{7}{2}\mathrm{sq}\mathrm{units} \end{gathered}$$

$$\begin{gathered} \mathrm{Consider}\mathrm{a}\mathrm{point}S\left(x,y_4\right)\mathrm{on}BC\mathrm{and}R\left(x,y_3\right)\mathrm{on}DC \\ \mathrm{For}\mathrm{a}\mathrm{vertical}\mathrm{strip}\mathrm{of}\mathrm{length}=\left|y_4-y_3\right|\mathrm{and}\mathrm{width}=dx,\mathrm{area}=\left|y_4-y_3\right|dx, \\ \mathrm{The}\mathrm{approximating}\mathrm{rectangle}\mathrm{moves}\mathrm{from}x=6\mathrm{to}x=8 \\ \Rightarrow \mathrm{area}\left(BDC\right)={\int }_6^8\left[\left(-x+12\right)-\left(\frac{3}{4}x-2\right)\right]dx \\ ={\int }_6^8\left(-\frac{7}{4}x+14\right)dx \\ ={\left[-\frac{7}{4}\times \frac{x^2}{2}+14x\right]}_6^8 \\ ={\left[-\frac{7x^2}{8}+14x\right]}_6^8 \\ =-\frac{7}{8}\left(64-36\right)+14\left(8-6\right) \\ =-\frac{7}{8}\times 28+28 \\ =\frac{28}{8}=\frac{7}{2} \\ \mathrm{Thus}\mathrm{required}\mathrm{area}\left(ABC\right)=\frac{7}{2}+\frac{7}{2}=7\mathrm{sq}.\mathrm{units} \end{gathered}$$