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प्रश्न
Using Integration, find the area of triangle whose vertices are (– 1, 1), (0, 5) and (3, 2).
उत्तर
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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