Question

Using the method of integration, find the area of the region bounded by the following lines:
(i) 3x - y - 3 = 0, 2x + y - 12 = 0, x- 2y - 1 = 0.
(ii) 3x - 2y + 1 = 0, 2x + 3y - 21 = 0 and x - 5y + 9 = 0

Answer 1

(ii) To find: area of the region bounded by the lines
3x − 2y + 1 = 0 ..(1)
2x + 3y − 21 = 0 ..(2)
x − 5y + 9 = 0 ..(3)

On solving (1) and (2), we get
x = 3 and y = 5

On solving (2) and (3), we get
x = 6 and y = 3

On solving (1) and (3), we get
x = 1 and y = 2




Equation of AB is
$$\begin{gathered} 3x-2y+1=0 \\ \Rightarrow 2y=3x+1 \\ \Rightarrow y=\frac{3x+1}{2}....\left(4\right) \end{gathered}$$


Equation of BC is
$$\begin{gathered} 2x+3y-21=0 \\ \Rightarrow 3y=-2x+21 \\ \Rightarrow y=\frac{-2x+21}{3}....\left(5\right) \end{gathered}$$


Equation of AC is
$$\begin{gathered} x-5y+9=0 \\ \Rightarrow 5y=x+9 \\ \Rightarrow y=\frac{x+9}{5}....\left(6\right) \end{gathered}$$


Now,
$$\begin{gathered} \mathrm{Area}\left(ABED\right)={\int }_1^{3}ydx \\ ={\int }_1^3\left(\frac{3x+1}{2}\right)dx \\ =\frac{1}{2}{\left(\frac{3x^2}{2}+x\right)}_1^3 \\ =\frac{1}{2}\left[\left(\frac{3\times 3^2}{2}+3\right)-\left(\frac{3\times 1^2}{2}+1\right)\right] \\ =\frac{1}{2}\left[\left(\frac{27}{2}+3\right)-\left(\frac{3}{2}+1\right)\right] \\ =\frac{1}{2}\left[\frac{27}{2}+3-\frac{3}{2}-1\right] \\ =\frac{1}{2}\left[\frac{24}{2}+2\right] \\ =\frac{1}{2}\left[12+2\right] \\ =\frac{1}{2}\left[14\right] \\ =7 \\ \mathrm{Thus},\mathrm{Area}\left(ABED\right)=7...\left(7\right) \\ \mathrm{Area}\left(BCFE\right)={\int }_3^{6}ydx \\ ={\int }_3^6\left(\frac{-2x+21}{3}\right)dx \\ =\frac{1}{3}{\left(\frac{-2x^2}{2}+21x\right)}_3^6 \\ =\frac{1}{3}\left[\left(\frac{-2\times 6^2}{2}+21\left(6\right)\right)-\left(\frac{-2\times 3^2}{2}+21\left(3\right)\right)\right] \\ =\frac{1}{3}\left[\left(-36+126\right)-\left(-9+63\right)\right] \\ =\frac{1}{3}\left[\left(90\right)-\left(54\right)\right] \\ =\frac{1}{3}\left[36\right] \\ =12 \\ \mathrm{Thus},\mathrm{Area}\left(BCFE\right)=12...\left(8\right) \end{gathered}$$


$$\begin{gathered} \mathrm{Area}\left(ACFD\right)={\int }_1^{6}ydx \\ ={\int }_1^6\left(\frac{x+9}{5}\right)dx \\ =\frac{1}{5}{\left(\frac{x^2}{2}+9x\right)}_1^6 \\ =\frac{1}{5}\left[\left(\frac{6^2}{2}+9\left(6\right)\right)-\left(\frac{1^2}{2}+9\left(1\right)\right)\right] \\ =\frac{1}{5}\left[\left(\frac{36}{2}+54\right)-\left(\frac{1}{2}+9\right)\right] \\ =\frac{1}{5}\left[\left(18+54\right)-\left(\frac{1+18}{2}\right)\right] \\ =\frac{1}{5}\left[72-\frac{19}{2}\right] \\ =\frac{1}{5}\left[\frac{144-19}{2}\right] \\ =\frac{1}{5}\left[\frac{125}{2}\right] \\ =\frac{25}{2} \\ \mathrm{Thus},\mathrm{Area}\left(ACFD\right)=\frac{25}{2}...\left(9\right) \end{gathered}$$


Area of ∆ABC = Area (ABED) + Area (BCFE) − Area (ACFD)
= 7 + 12 − $\frac{25}{2}$
$$\begin{gathered} =19-\frac{25}{2} \\ =\frac{38-25}{2} \\ =\frac{13}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$


Hence, area of the required region is $\frac{13}{2}$ sq. units.