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Question
Compute the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7.
Solution
Given that: x + 2y = 2 .....(i)
y – x = 1 ......(ii)
And 2x + y = 7 ......(iii)
| x | 0 | 2 |
| y | 1 | 0 |
| x | 0 | –1 |
| y | 1 | 0 |
| x | 0 | `7/2` |
| y | 7 | 0 |
Solving equations (ii) and (iii)
We get y = 1 + x
∴ 2x + 1 + x = 7
3x = 6
⇒ x = 2
∴ y = 1 + 2
= 3
Coordinates of B = (2, 3)
Solving equations (i) and (iii)
We get x + 2y = 2
∴ x = 2 – 2y
2x + y = 7
2(2 – 2y) + y = 7
⇒ 4 – 4y + y = 7
⇒ –3y = 3
∴ y = –1 and x = 4
∴ Coordinates of C = (4, – 1) and coordinates of A = (0, 1).
Taking the limits on y-axis, we get
`int_(-1)^3 x_"BC" "d"y - int_(-1)^1 x_"AC" "d"y - int_1^3 x_"AB" "d"y`
= `int_(-1)^3 (7 - y)/2 "d"y - int_(-1)^1 (2 - 2y) "d"y - int_1^3 (y - 1) "d"y`
= `1/2 [7y - y^2/2]_-1^2 - 2[y - y^2/2]_-1^1 - [y^2/2 - y]_1^3`
= `1/2[(21 - 9/2) - (7 - 1/2)] - 2[(1 - 1/2) - (-1 - 1/2)] - [(9/2 - 3) - (1/2 - 1)]`
= `1/2[33/2 + 15/2] - 2[1/2 + 3/2] - [3/2 + 1/2]`
= `1/2 xx 24 - 2 xx 2 - 2`
⇒ 12 – 4 – 2 = 6 sq.units
Hence, the required area = 6 sq.units
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