Question

Find the area bounded by the lines y = 4x + 5, y = 5 − x and 4y = x + 5.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ y=4x+5.....\left(1\right) \\ y=5-x.....\left(2\right) \\ 4y=x+5.....\left(3\right) \end{gathered}$$
All the three equations represent equations of straight lines
The points of intersection is obtained by solving simultaneous equations
$$\begin{gathered} \mathrm{From}\left(1\right)\mathrm{and}\left(2\right) \\ 4x+5=5-x \\ \Rightarrow 5x=0 \\ \Rightarrow x=0 \\ \Rightarrow y=5 \\ \mathrm{Thus}A\left(0,5\right)\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\left(1\right)\mathrm{a}\mathrm{n}\mathrm{d}\left(2\right) \\ \mathrm{From}\left(2\right)\mathrm{and}\left(3\right) \\ 4\left(5-x\right)=x+5 \\ \Rightarrow 5x=15 \\ \Rightarrow x=3 \\ \Rightarrow y=2 \\ \mathrm{Thus}B\left(3,2\right)\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\left(2\right)\mathrm{a}\mathrm{n}\mathrm{d}\left(3\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(3\right) \\ 4\left(4x+5\right)=x+5 \\ \Rightarrow 15x=-15 \\ \Rightarrow x=-1 \\ \Rightarrow y=1 \\ \mathrm{Thus}C\left(-1,1\right)\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\left(1\right)\mathrm{a}\mathrm{n}\mathrm{d}\left(3\right) \\ \mathrm{Area}\left(\mathit{A}\mathit{B}\mathit{C}\right)=\mathrm{area}\left(\mathit{A}\mathit{B}\mathit{P}\right)+\mathrm{area}\left(\mathit{P}\mathit{A}\mathit{B}\right) \\ ={\int }_{-1}^0\left[\left(4x+5\right)-\left(\frac{x+5}{4}\right)\right]dx+{\int }_0^3\left[\left(5-x\right)-\left(\frac{x+5}{4}\right)\right]dx \\ ={\int }_{-1}^0\left(\frac{15}{4}x+\frac{15}{4}\right)dx+{\int }_0^3\left(\frac{15}{4}-\frac{5}{4}x\right)dx \\ =\frac{15}{4}{\left[\frac{x^2}{2}+x\right]}_{-1}^0+\frac{5}{4}{\left[3x-\frac{x^2}{2}\right]}_0^3 \\ =\frac{15}{4}\left(-\frac{1}{2}+1\right)+\frac{5}{4}\left(9-\frac{9}{2}\right) \\ =\frac{15}{8}+\left(\frac{5}{4}\times \frac{9}{2}\right) \\ =\frac{15}{8}+\frac{45}{8} \\ =\frac{60}{8} \\ =\frac{15}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$