Using integration find the area of region bounded by the triangle whose vertices are (1, 0), (2, 2) and (3, 1).

Explanation

Equation of AB
`y-0=(2-0)/(2-1)(x-1)`
` implies y=2x-2`
Equation of BC
`y-2=(1-2)/(3-2)(x-2)impliesy=4-x`

Equation of AC
`y-0=(1-0)/(3-1)(x-1)impliesy=(1)/(2)(x-1)`
Now area of `triangle ABC`
`="Area of "triangle ABD +"Area of " square DBCE - "Area of " triangle ACE`
`int_(1)^(2)(2x-2)dx+int_(2)^(3)(4-x)dx-int_(1)^(3)(1)/(2)(x-1)dx`
`=[x^(2)-2x]_(1)^(2)+[4x-(x^(2))/(2)]_(2)^(3)-(1)/(2)[(x^(2))/(2)-x]_(1)^(3)`
`=[(4-4)-(1-2)]+[(12-(9)/(2))-(8-2)]-(1)/(2)[((9)/(2)-3)-((1)/(2)-1)]`
`=1+(15)/(2)-6-(3)/(4)-(1)/(4)=(3)/(2)sq.` units.