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

Given lines are
$3x-y-3=0$ --- (1)
$2x+y-12=0$--- (2)
$x-2y-1=0$ ---- (3)
For intersecting point of (1) and (2)
(1) + (2) $\Rightarrow 3x-y-3+2x+y-12=0$
$\Rightarrow 5x-15=0\Rightarrow x=3$
Putting x = 3 in (1), we get
$9-y-3=0$
$y=6$
Intersecting point of (1) and (2) is (3, 6)
For intersecting point of (2) and (3)
$\left(2\right)-2\times \left(3\right)\Rightarrow 2x+y-12-2x+4y+2=0$
$\Rightarrow 5y=10=0$
$\Rightarrow y=2$
Putting y= 2 in (2), we get
$2x+2-12=0$
$x=5$
Intersecting point of (2) and (3) is (5, 2)
For intersecting point of (1)and (3)
$\left(1\right)-3\times \left(3\right)\Rightarrow 3x-y-3-3x+6y+3=0$
$\Rightarrow 5y=0$
$\Rightarrow y=0$
Putting y = 0 in (1), we get
3x - 3 = 0
x = 1
Intersecting point (1) and (3) is (1, 0)
Shaded region is required region
$\therefore$ Required area =${\int }_1^3(3x-3)dx+{\int }_3^5(-2x+12)dx-{\int }_1^5\frac{x-1}{2}dx$
$=3{\int }_1^{3}xdx-3{\int }_1^{3}dx-2{\int }_3^{5}xdx+12{\int }_3^{5}dx-\frac{1}{2}{\int }_1^{5}xdx+\frac{1}{2}{\int }_1^{5}dx$
$=3{\left[\frac{x^2}{2}\right]}_1^3-3{\left[x\right]}_1^3-2{\left[\frac{x^2}{2}\right]}_3^5+12{\left[x\right]}_3^5-\frac{1}{2}{\left[\frac{x^2}{2}\right]}_1^5+\frac{1}{2}{\left[x\right]}_1^5$
$=3(9-1)-3(3-1)-(25-9)+12(5-3)-\frac{1}{4}(25-1)+\frac{1}{2}(5-1)$
$=12-6-16+24-6+2$
$=10$ sq.unit
Therefore the area is 10 sq.units.