Question

A farmer has a field bounded by three lines x + 2y = 2, y – x = 1, 2x + y = 7. Using integration, find the area of the region bounded by these lines.

Answer 1

Lines are

x + 2y = 2   ...(i)

y – x = 1   ...(ii)

And 2x + y = 7   ...(iii)

Solving equations (i) and (ii),

x + 2y = 2

–x + y = 1

On adding, we get

3y = 3

⇒ y = 1

Substitute y = 1 in equation (ii),

1 – x = 1

⇒ x = 0

The intersection point is A(0, 1).

Solving equations (ii) and (iii),

–x + y = 1

2x + y = 7

On subtracting, we get

–3x = – 6

x = 2

Substitute x = 2 in equation (ii),

–2 + y = 1

⇒ y = 3

The intersection point is B(2, 3).

Solving equations (i) and (iii),

x + 2y = 2

2x + y = 7   ...(Multiply by 2)

On subtracting, we get

–3x = –12

⇒ x = 4

Substitute x = 4 in equation (i)

4 + 2y = 2

⇒ y = –1

The intersection point is C(4, –1).

The vertices of the triangle are (0, 1), (4, –1) and (2, 3).

By the above figure.

Required area

= `int_0^2 "AB line"  dx + int_2^4 "BC line"  dx - int_0^4 "AC line"  dx`

= `int_0^2 (x + 1) dx + int_2^4 (7 - 2x) dx - int_0^4 ((2 - x)/2) dx`

= `(x^2/2 + x)_0^2 + (7x - x^2)_2^4 - 1/2 (2x - x^2/2)_0^4`

= `4/2 + 2 + 28 - 16 - 14 + 4 - 1/2 (8 - 8 - 0 + 0)`

= 6 sq. units

Area of required region = 6 sq. units.