Using the method of integration find the area of the region bounded by lines:`2x + y = 4, 3x- 2y = 6` and `x - 3y + 5 = 0`

Explanation

Equation of given lines are,
`2x+y=4 " " ` …(1)
`3x-2y=6 " " `…(2)
`x-3y+5=0 " " ` …(3)
From equations (1) and (2),
`4-2x=(3x-6)/(2)`

`implies 8-4x=3x-6`
`implies 7x=14 implies x=2`
From equation (1),
`y=4-2x=4-2xx2=0`
` :. ` Equation (1) and (2) intersect each other at point (2, 0).
From equations (2) and (3),
`(3x-6)/(2)=(x+5)/(3)implies 9x-18=2x+10`
` implies 7x=28 implies x=4`
From equation (2), `y=(3xx4-6)/(2)=3`
` :. ` Equation (2) and (3) intersect each other at point (4, 3).
From equations (1) and (2), `4-2x=(x+5)/(3)`
`implies 7x=7 implies x=1`
From equation (1), `y=4-2x=4-2xx1=2`
Equation (1) and (3) intersect each other at point (1, 2).
` :. ` Required area `=int_(1)^(4)((x+5)/(3))dx - int_(1)^(2)(4-2x)dx -int_(2)^(4)((3x-6)/2)dx`
`=(1)/(3)[((x+5)^(2))/(2)]_(1)^(4)-[4x-x^(2)]_(1)^(2)-(1)/(2)[(3x^(2))/(2)-6x]_(2)^(4)`
`=(1)/(6)[9^(2)-6^(2)]-{(8-4)-(4-1)}-(1)/(2)[((3xx4^(2))/(2)-6xx4)-(3xx(2^(2))/(2)-6xx2)]`
`=(45)/(6)-1-(1)/(2)(0-6+12)`
`=(15)/(2)-1-3=(7)/(2)` sq. units.