Question:
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).
Answer:
Given vertices of triangle are: A(2, 0), B (4, 5) and C (6, 3)
Now equation of line AB = (y - 0)/(5 -0) = (x-2)/(4-2)
=> y/5 = (x-2)/2
=> y = 5/2 *(x-2) ...............1
Equation of line BC = (y - 5)/(3 -5) = (x-4)/(6-4)
=> (y - 5)/-2 = (x-4)/2
=> -(y - 5) = (x-4)
=> -y + 5 = x-4
=> x + y = 9
=> y = -x + 9 .......2
Equation of line CA = (y - 3)/(0 -3) = (x-2)/(2-6)
=> (y - 3)/-3 = (x-2)/-4
=> 4(y - 3) = 3(x-2)
=> y-3 = 3/4 *(x-2)
=> y = 3/4 *(x-2) + 3 ........3
Now area enclosed by these three lines is
A = 4∫2 5/2 (x-2)dx + 1∫0 (-x + 9)dx + 2∫6 3/4 (x-2)dx
=> A = 5/2 *[x2 /2 - 2x 2]4 + [-x2 /2 + 9x 0]1 - 3/4*[x2 /2 - 2x 2]6
=> A = 5/2 * [8 - 8 -2+4] + [-1/2 + 9 - 0] - 3/4 *[18-12-2+4]
=> A = 5/2 * 2 + 17/2 - 3/4 *8
=> A = 5 + 17/2 - 6
=> A = 17/2 - 1
=> A = 15/2
=> A = 7.5
So area is 7.5 unit2
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