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Let O be the origin and A, B be the two points having coordinates (0, 4) and (6, 0) respectively. If a point P moves in such a way that the area of the `Delta OPA` is always twice the area of `Delta POB`, then P lies on

A

`y^(2) = 9x^(2)`

B

`x^(2) = - 9y^(2)`

C

`x^(2) - 9y^(2) = 0`

D

None of these

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To solve the problem, we need to find the locus of the point \( P(x, y) \) such that the area of triangle \( OPA \) is always twice the area of triangle \( POB \). ### Step-by-Step Solution: 1. **Identify the Points**: - Let \( O(0, 0) \) be the origin. - Let \( A(0, 4) \) and \( B(6, 0) \) be the given points. 2. **Area of Triangle OPA**: The area of triangle \( OPA \) can be calculated using the formula: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] For triangle \( OPA \): - \( O(0, 0) \) corresponds to \( (x_1, y_1) = (0, 0) \) - \( P(x, y) \) corresponds to \( (x_2, y_2) = (x, y) \) - \( A(0, 4) \) corresponds to \( (x_3, y_3) = (0, 4) \) Substituting these coordinates into the formula gives: \[ \text{Area}_{OPA} = \frac{1}{2} \left| 0(y - 4) + x(4 - 0) + 0(0 - y) \right| = \frac{1}{2} \left| 4x \right| = 2x \] 3. **Area of Triangle POB**: For triangle \( POB \): - \( P(x, y) \) corresponds to \( (x_1, y_1) = (x, y) \) - \( O(0, 0) \) corresponds to \( (x_2, y_2) = (0, 0) \) - \( B(6, 0) \) corresponds to \( (x_3, y_3) = (6, 0) \) Substituting these coordinates gives: \[ \text{Area}_{POB} = \frac{1}{2} \left| x(0 - 0) + 0(0 - y) + 6(y - 0) \right| = \frac{1}{2} \left| 6y \right| = 3y \] 4. **Setting Up the Condition**: According to the problem, the area of triangle \( OPA \) is twice the area of triangle \( POB \): \[ 2x = 2 \times 3y \] Simplifying this gives: \[ 2x = 6y \quad \Rightarrow \quad x = 3y \] 5. **Finding the Locus**: Rearranging the equation \( x = 3y \) gives: \[ x - 3y = 0 \] This is the equation of the line on which point \( P \) lies. ### Final Answer: The locus of point \( P \) is given by the equation: \[ x - 3y = 0 \]
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