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Question
Find the area bounded by the lines y = 4x + 5, y = 5 – x and 4y = x + 5.
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Solution
Given that y = 4x + 5 .....(i)
y = 5 – x ......(ii)
And 4y = x + 5 ......(iii)
| x | 0 | `-5/4` |
| y | 5 | 0 |
| x | 0 | 5 |
| y | 5 | 0 |
| x | 0 | –5 |
| y | `5/4` | 0 |
Solving equations (i) and (ii)
We get 4x + 5 = 5 – x
⇒ x = 0 and y = 5
∴ Coordinates of A = (0, 5)
Solving equations (ii) and (iii)
y = 5 – x
4y = x + 5
5y = 10
∴ y = 2 and x = 3
∴ Coordinates of B = (3, 2)
Solving equations (i) and (iii)
y = 4x + 5
4y = x + 5
⇒ 4(4x + 5) = x + 5
⇒ 16x + 20 = x + 5
⇒ 15x = – 15
∴ x = –1 and y = 1
∴ Coordinates of C = (–1, 1).
∴ Area of required regions = `int_(-1)^0 y_"AC" "d"x + int_0^3 y_"AB" "d"x - int_(-1)^3 y_"CB" "d"x`
= `int_(-1)^0 (4x + 5) "d"x + int_0^3 (5 - x) "d"x - int_(-1)^3 (x + 5)/4 "d"x`
= `[4 x^2/2 + 5x]_-1^0 + [5x - x^2/2]_0^3 - 1/4[x^2/2 + 5x]_-1^3`
= `[(0 + 0) - (2 - 5)] + [(15 - 9/2) - (0 - 0)] - 1/4[(9/2 + 15) - (1/2 - 5)]`
= `3 + 21/2 - 1/4[39/2 + 9/2]`
= `3 + 21/2 - 1/4 xx 24`
⇒ `3 + 21/2 - 6`
= `15/2` sq.units
Hence, the required area = `15/2` sq.units
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