Question

Using integration find the area of the region bounded by the triangle whose vertices are
(i) (-1, 2), (1, 5) and (3, 4) (ii) (-2, 1), (0, 4) and (2, 3)
(iii) (2,5), (4,7) and (6, 2)

Answer 1

(iii) To find: area of the region bounded by the triangle whose vertices are (2, 5), (4, 7) and (6, 2).



We know,
Equation of a line joining two points (x₁, y₁) and (x₂, y₂) is

$y-y_1=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)$ ...(1)

Equation of AB is
$$\begin{gathered} y-5=\frac{7-5}{4-2}\left(x-2\right) \\ \Rightarrow y-5=\frac{2}{2}\left(x-2\right) \\ \Rightarrow y-5=1\left(x-2\right) \\ \Rightarrow y=x-2+5 \\ \Rightarrow y=x+3....\left(2\right) \end{gathered}$$


Equation of BC is
$$\begin{gathered} y-2=\frac{7-2}{4-6}\left(x-6\right) \\ \Rightarrow y-2=\frac{5}{-2}\left(x-6\right) \\ \Rightarrow -2\left(y-2\right)=5\left(x-6\right) \\ \Rightarrow -2y+4=5x-30 \\ \Rightarrow -2y=5x-30-4 \\ \Rightarrow -2y=5x-34 \\ \Rightarrow 2y=-5x+34 \\ \Rightarrow y=\frac{-5x+34}{2}....\left(3\right) \end{gathered}$$


Equation of AC is
$$\begin{gathered} y-5=\frac{2-5}{6-2}\left(x-2\right) \\ \Rightarrow y-5=\frac{-3}{4}\left(x-2\right) \\ \Rightarrow 4\left(y-5\right)=-3\left(x-2\right) \\ \Rightarrow 4y-20=-3x+6 \\ \Rightarrow 4y=-3x+6+20 \\ \Rightarrow 4y=-3x+26 \\ \Rightarrow y=\frac{-3x+26}{4}....\left(4\right) \end{gathered}$$


Now,
$$\begin{gathered} \mathrm{Area}\left(ABED\right)={\int }_2^{4}ydx \\ ={\int }_2^4\left(x+3\right)dx \\ ={\left(\frac{x^2}{2}+3x\right)}_2^4 \\ =\left(\frac{4^2}{2}+3\left(4\right)\right)-\left(\frac{2^2}{2}+3\left(2\right)\right) \\ =\left(\frac{16}{2}+12\right)-\left(\frac{4}{2}+6\right) \\ =\left(8+12\right)-\left(2+6\right) \\ =20-8 \\ =12 \\ \mathrm{Thus},\mathrm{Area}\left(ABED\right)=12...\left(5\right) \\ \mathrm{Area}\left(BCFE\right)={\int }_4^{6}ydx \\ ={\int }_4^6\left(\frac{-5x+34}{2}\right)dx \\ =\frac{1}{2}{\left(\frac{-5x^2}{2}+34x\right)}_4^6 \\ =\frac{1}{2}\left[\left(\frac{-5\times 6^2}{2}+34\left(6\right)\right)-\left(\frac{-5\times 4^2}{2}+34\left(4\right)\right)\right] \\ =\frac{1}{2}\left[\left(\frac{-5\times 36}{2}+204\right)-\left(\frac{-5\times 16}{2}+136\right)\right] \\ =\frac{1}{2}\left[\left(-5\times 18+204\right)-\left(-5\times 8+136\right)\right] \\ =\frac{1}{2}\left[\left(-90+204\right)-\left(-40+136\right)\right] \\ =\frac{1}{2}\left[\left(114\right)-\left(96\right)\right] \\ =\frac{1}{2}\left[18\right] \\ =9 \\ \mathrm{Thus},\mathrm{Area}\left(BCFE\right)=9...\left(6\right) \end{gathered}$$


$$\begin{gathered} \mathrm{Area}\left(ACFD\right)={\int }_2^{6}ydx \\ ={\int }_2^6\left(\frac{-3x+26}{4}\right)dx \\ =\frac{1}{4}{\left(\frac{-3x^2}{2}+26x\right)}_2^6 \\ =\frac{1}{4}\left[\left(\frac{-3\times 6^2}{2}+26\left(6\right)\right)-\left(\frac{-3\times 2^2}{2}+26\left(2\right)\right)\right] \\ =\frac{1}{4}\left[\left(\frac{-3\times 36}{2}+156\right)-\left(\frac{-3\times 4}{2}+52\right)\right] \\ =\frac{1}{4}\left[\left(-3\times 18+156\right)-\left(-3\times 2+52\right)\right] \\ =\frac{1}{4}\left[\left(-54+156\right)-\left(-6+52\right)\right] \\ =\frac{1}{4}\left[\left(102\right)-\left(46\right)\right] \\ =\frac{1}{4}\left[56\right] \\ =14 \\ \mathrm{Thus},\mathrm{Area}\left(ACFD\right)=14...\left(7\right) \end{gathered}$$


Area of ∆ABC = Area (ABED) + Area (BCFE) − Area (ACFD)
= 12 + 9 − 14
= 21 − 14
= 7 sq. units


Hence, area of the required region is 7 sq. units.