Question

Find the area of region bounded by the triangle whose vertices are (–1, 1), (0, 5) and (3, 2), using integration.

Answer 1

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