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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.
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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