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 joining A and B is `(y-0)/(x-2) = (5-0)/(4-2) => y = (5x)/(2) -5`
Equation of line joining B and C is `(y-5)/(x-4) = (5-3)/(4-6) => y = 9-x`
Equation of line joining A and C is `(y-0)/(x-2) = (3-0)/(6-2) => y = (3x)/(4) -3/2`
ar(ABC)=ar(ABL)+ar(LBCM)-ar(ACM)
Now
ar(ABL)=` int_2^4 ( (5x)/(2) -5) dx = [ (5x^2)/4 -5x]_2^4 = (20-20)-(5-10) = 5` units
ar(LBCM)=` int_4^6 ( 9-x) dx = [ 9x -x^2/2]_4^6 = (54-18)-(36-8) = 8` units
ar(ACM) = ` int_2^6 ( (3x)/4-3/2) dx = [(3x^2)/(8) -(3x)/2]_2^6 = ( (27)/2 -9)-(3/(2)- 3) = 6` units
Hence
ar(ABC) = `8+5-6 = 7` units