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Question
Using Integration, find the area of triangle whose vertices are (– 1, 1), (0, 5) and (3, 2).
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Solution
The vertices of the triangle are P(– 1, 1), Q(0, 5) and R(3, 2).
Equation of line PQ
y – 1 = `((5 - 1))/((0 + 1)) (x - 1)`
`\implies` y – 1 = 4(x + 1)
`\implies` y – 1 = 4x + 4
`\implies` y = 4x + 5 ...(i)
Equation of line QR
y – 5 = `((2 - 5))/((3 + 0)) (x - 0)`
`\implies` y – 5 = `- 3/3 x`
`\implies` y = – x + 5 ...(ii)
Equation of line PR
y – 1 = `((2 - 1))/((3 + 1)) (x + 1)`
`\implies` y – 1 = `1/4 (x + 1)`
`\implies` 4y – 4 = x + 1
`\implies` 4y = x + 5
`\implies` y = `x/4 + 5/4` ...(iii)
Area of ΔPQR = (Area under PQ) + (Area under QR) – (Area under PR)
= `int_-1^0 (4x + 5)dx + int_0^3 (-x + 5)dx - int_-1^3 (x/4 + 5/4)dx`
= `[(4x^2)/2 + 5x]_-1^0 + [(-x^2)/2 + 5x]_0^3 - [x^2/8 + (5x)/4]_-1^3`
= `[-(2 - 5)] + [((-9)/2 + 15)] - [(9/8 + 15/4) - (1/8 - 5/4)]`
= `3 + [((-9 + 30)/2)] - [((9 + 30)/8) - ((1 - 10)/8)]`
= `3 + [(21/2)] - [39/8 + 9/8]`
= `3 + (21/2) - 48/8`
= `3 + 21/2 - 6`
= `15/2` sq. unit.
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