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

The equation of side AB is
`(y-0)/(x-2)=(5-0)/(4-2) rArr y-5/2(x-2)`

The equation of side BC is
`(y-5)/(x-4)=(3-5)/(6-4)rArr y=-x+9`
The equation of side AC is
`(y-0)/(x-2)=(3-0)/(6-2)rArr y=(3)/(4)( x-2)`
Draw perpendiculars BL and CM on the x-axies
`therefore` area of `Delta ABC`
`=ar(Delta ALB)+ar(trap. BLMC )-ar(Delta AMC)`
`=underset(4)overset(2) (int) y_(AB) dx + overset(6)underset(4)(int) (9-x) d x - (3)/(4) overset(6)underset(2)(int) y_(BC) dx - overset(6)underset(2) (int) y_(AC) dx`
` = (5)/(2) overset(4)underset(2) (int) (x-2) dx + overset(6)underset(4)(int) (9-x) dx - (3)/(4) overset(6)underset(2)(int) (x-2) dx`
`= (5)/(2) [(x^(2))/(2) - 2x)]_(2)^(4) + [9x - (x^(2))/(2)]_(4)^(6) - (3)/(4) [(x^(2))/(2) - 2x ]_(2)^(6)`
`= (5)/(2) [0-(-2)] + (36-28) - (3)/(4) [6-(2-)]`
= `(5 + 8 - 6)` sq units = 7 sq units