Question

Using the method of integration, find the area of the region bounded by the following lines:
3x − y − 3 = 0, 2x + y − 12 = 0, x − 2y − 1 = 0.

Answer 1

We have,

$$\begin{gathered} 3x-y-3=0\cdots \left(1\right) \\ 2x+y-12=0...\left(2\right) \\ x-2y-1=0...\left(3\right) \end{gathered}$$


$$\begin{gathered} \mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}, \\ 5x-15=0 \\ \Rightarrow x=3 \\ \therefore y=6 \\ B(3,6)\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\left(1\right)\mathrm{and}\left(2\right) \\ \mathrm{Solving}\left(1\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get}, \\ 5x=5 \\ \Rightarrow x=1 \\ \therefore y=0 \\ A\left(1,0\right)\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\left(1\right)\mathrm{and}\left(3\right) \\ \mathrm{Solving}\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get}, \\ 5x=25 \\ \Rightarrow x=5 \\ \therefore y=2 \\ C\left(5,2\right)\mathrm{is}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\left(2\right)\mathrm{and}\left(3\right) \\ \mathrm{Now}, \\ \mathrm{Area}ABC=\left\{\mathrm{area}\mathrm{bound}\mathrm{by}\left(1\right)\mathrm{between}x=1\mathrm{and}x=3\right\}+\left\{\mathrm{area}\mathrm{bound}\mathrm{by}\left(2\right)\mathrm{between}x=3\mathrm{and}x=5\right\}-\left\{\mathrm{area}\mathrm{bound}\mathrm{by}\left(3\right)\mathrm{between}x=1\mathrm{and}x=5\right\} \\ ={\int }_1^3\left(3x-3\right)dx+{\int }_3^5\left(12-2x\right)dx-{\int }_1^5\frac{\left(x-1\right)}{2}dx \\ ={\left[3\times \frac{x^2}{2}-3x\right]}_1^3+{\left[12x-2\times \frac{x^2}{2}\right]}_3^5-\frac{1}{2}{\left[\frac{x^2}{2}-x\right]}_1^5 \\ =3{\left[\frac{x^2}{2}-x\right]}_1^3+{\left[12x-x^2\right]}_3^5-\frac{1}{2}{\left[\frac{x^2}{2}-x\right]}_1^5 \\ =3\left(\frac{9}{2}-3\right)-3\left(\frac{1}{2}-1\right)+\left(12-25\right)-\left(12-9\right)-\frac{1}{2}\left[\left(\frac{25}{2}-1\right)-\left(\frac{1}{2}-1\right)\right] \\ =6+8-4 \\ =10\mathrm{sq}\mathrm{units} \end{gathered}$$