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Question
Find the area of region bounded by the triangle whose vertices are (–1, 1), (0, 5) and (3, 2), using integration.
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Solution
The coordinates of the vertices of ΔABC are given by A(–1, 1), B(0, 5) and C(3, 2).
Equation of AB is y – 1 = `(5 - 1)/(0 + 1) (x + 1)`
⇒ y – 1 = 4x + 4
∴ y = 4x + 4 + 1
⇒ y = 4x + 5 .....(i)
Equation of BC is y – 5 = `(2 - 5)/(3 - 0) (x - 0)`
⇒ y – 5 = –x
∴ y = 5 – x ......(ii)
Equation of CA is y – 1 = `(2 - 1)/(3 + 1) (x + 1)`
⇒ y – 1 = `1/4x + 1/4`
⇒ y = `1/4x + 1/4 + 1`
∴ y = `1/4x + 5/4`
= `1/4 (5 + x)`
Area of ΔABC = `int_(-1)^0 (4x + 5) "d"x + int_0^3 (5 - x) "d"x - int_(-1)^3 1/4(5 + x)"d"x`
= `4/2 [x^2]_-1^0 + 5[x]_-1^0 + 5[x]_0^3 - 1/2 [x^2]_0^3 - 1/4 [5x + x^2/2]_-1^3`
= `2(0 - 1) + 5(0 + 1) + 5(3 - 0) - 1/2 (9 - 0) - 1/4[(15 + 9/2) - (-5 + 1/2)]`
= `-2 + 5 + 15 - 9/2 - 1/4 (39/2 + 9/2)`
= `18 - 9/2 - 1/4 xx 48/2`
= `18 - 9/2 - 6`
= `12 - 9/2`
= `15/2` sq.units
Hence, the required area = `15/2` sq.units
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