Question

Using integration, find the area of the region bounded by the triangle whose vertices are (−1, 2), (1, 5) and (3, 4). [CBSE 2014]

Answer 1

Let ABC be the triangle with vertices A(−1, 2), B(1, 5) and C(3, 4).



Equation of AB is

$$\begin{gathered} y-5=\left(\frac{2-5}{-1-1}\right)\left(x-1\right) \\ \Rightarrow y-5=\frac{3}{2}\left(x-1\right) \\ \Rightarrow y=\frac{3}{2}x+5-\frac{3}{2}=\frac{3x+7}{2} \end{gathered}$$

Equation of BC is

$$\begin{gathered} y-4=\left(\frac{5-4}{1-3}\right)\left(x-3\right) \\ \Rightarrow y-4=-\frac{1}{2}\left(x-3\right) \\ \Rightarrow y=-\frac{1}{2}x+4+\frac{3}{2}=\frac{-x+11}{2} \end{gathered}$$

Equation of CA is

$$\begin{gathered} y-2=\left(\frac{4-2}{3+1}\right)\left(x+1\right) \\ \Rightarrow y-2=\frac{1}{2}\left(x+1\right) \\ \Rightarrow y=\frac{1}{2}x+2+\frac{1}{2}=\frac{x+5}{2} \end{gathered}$$

∴ Required area = Area of the shaded region

= Area of the region ABEFA + Area of the region BCDEB − Area of the region ACDFA

$$\begin{gathered} ={\int }_{-1}^1{y}_{\mathrm{AB}}dx+{\int }_1^3{y}_{\mathrm{BC}}dx-{\int }_{-1}^3{y}_{\mathrm{CA}}dx \\ ={\int }_{-1}^1\left(\frac{3x+7}{2}\right)dx+{\int }_1^3\left(\frac{-x+11}{2}\right)dx-{\int }_{-1}^3\left(\frac{x+5}{2}\right)dx \\ =\frac{1}{2}\times {\overline{)\frac{{\left(3x+7\right)}^2}{2\times 3}}}_{-1}^1+\frac{1}{2}\times {\overline{)\frac{{\left(-x+11\right)}^2}{2\times \left(-1\right)}}}_1^3-\frac{1}{2}\times {\overline{)\frac{{\left(x+5\right)}^2}{2}}}_{-1}^3 \\ =\frac{1}{12}\left(100-16\right)-\frac{1}{4}\left(64-100\right)-\frac{1}{4}\left(64-16\right) \\ =\frac{84}{12}+\frac{36}{4}-\frac{48}{4} \\ =7+9-12 \\ =4\mathrm{square}\mathrm{units} \end{gathered}$$