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The area of the triangle formed by the graph of 3x + 4y = 12, x axis and y-axis (in sq. units) is
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- 4
- 12
- 6
- 8
- 4
Correct Option: C
x – axis Þ y = 0, putting in equation 3x + 4y = 12 3x = 12 ⇒ x = 4
⇒ Co-ordinates of point of intersection on x-axis = (4, 0)
Putting x = 0 in the equation 3x + 4y = 12 4y = 12 ⇒ y = 3
∴ Co-ordinates of point of intersection on y – axis = (0, 3)
∴ OA = 4
OB = 3
∴ Area of ∴OAB = | × OA × OB | × 4 × 3 = 6 sq. units | ||
2 | 2 |