Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A(2, 0), B (4, 5) and C (6, 3).

Explanation

Equation of line AB,
`y-0=(5-0)/(4-2)(x-2)`

`implies y=(5)/(2)x-5`
Equation of BC,
`y-5=(3-5)/(6-4)(x-4)`
`implies y= -x+9`
Equation of CA, `y-0=(3-0)/(6-2)(x-2)`
` implies y=(3)/(4)x-(3)/(2)`
`ar(triangle ABC)=ar(ABCDA)+ar(DBCED) - ar(ACEA)`
`=int_(2)^(4)((5)/(2)x-5)dx+int_(4)^(6)(9-x)dx-int_(2)^(6)((3)/(4)x-(3)/(2))dx`
`=[(5)/(2)x^(2)-5x]_(2)^(4)+[9x-(x^(2))/(2)]_(4)^(6)-[(3)/(8)x-(3)/(2)x]_(2)^(6)`
`=[(20-20)-(5-10)]+[(54-18)-(36-8)]-[((27)/(2)-9)-((3)/(2)-3)]`
`=5+36-28-(9)/(2)-(3)/(2)`
`=7` sq. units.