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The area in sq. unit. of the triangle formed by the graphs of x = 4, y = 3 and 3x + 4y = 12 is
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- 12
- 8
- 10
- 6
- 12
Correct Option: D
x = 4, a straight line parallel to y – axis.
y = 3, a straight line parallel to x – axis.
Putting x = 0 in 3x + 4y = 12, 3 × 0 + 4y = 12,
| ⇒ 4y = 12 ⇒ y = | = 4 | |
| 3 |
∴ Point of intersection on y – axis = (0, 4)
Again, putting y = 0 in 3x + 4y = 12, 3x + 4 × 0 = 12
| ⇒ 3x = 12 ⇒ x = | = 4 | |
| 3 |
∴ Point of intersection onx - axis = (4, 0)
Area of ∎ OACB = OA × OB = 4 × 3 = 12 sq. units
| Area of ∆ OAB = | × OA × OB = | × 4 × 3 = 6 sq. units | ||
| 2 | 2 |
∴ Area of ∆ABC = 12 – 6 = 6 sq. units
