Let P and Q be any two points on the lines represented by `2x-3y = 0` and `2x + 3y = 0` respectively. If the area of triangle OPQ (where O is origin) is 5, then which of the following is not the possible equation of the locus of mid-point of PO?

Options

A. `4x^(2) - 9y^(2) +30 = 0`
B. `4x^(2) - 9y^(2) - 30 = 0`
C. `9x^(2) - 4y^(2) - 30 = 0`
D. none of these

Explanation

To solve the problem, we need to find the locus of the midpoint of the line segment OP, where P is a point on the line \(2x - 3y = 0\) and Q is a point on the line \(2x + 3y = 0\), given that the area of triangle OPQ is 5.

### Step-by-Step Solution:

1. **Identify Points P and Q**:
- The line \(2x - 3y = 0\) can be rewritten as \(y = \frac{2}{3}x\). Let \(P\) be a point on this line, represented as \(P(h, \frac{2h}{3})\).
- The line \(2x + 3y = 0\) can be rewritten as \(y = -\frac{2}{3}x\). Let \(Q\) be a point on this line, represented as \(Q(k, -\frac{2k}{3})\).

2. **Calculate the Area of Triangle OPQ**:
- The area of triangle formed by points \(O(0,0)\), \(P(h, \frac{2h}{3})\), and \(Q(k, -\frac{2k}{3})\) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
- Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| 0\left(\frac{2h}{3} + \frac{2k}{3}\right) + h\left(-\frac{2k}{3} - 0\right) + k\left(0 - \frac{2h}{3}\right) \right| \]
- This simplifies to:
\[ \text{Area} = \frac{1}{2} \left| -\frac{2hk}{3} + \frac{2hk}{3} \right| = \frac{1}{2} \left| -\frac{4hk}{3} \right| = \frac{2hk}{3} \]
- Setting this equal to 5 (as given):
\[ \frac{2hk}{3} = 5 \implies hk = \frac{15}{2} \]

3. **Find the Midpoint M of OP**:
- The midpoint \(M\) of segment \(OP\) is given by:
\[ M\left(\frac{h}{2}, \frac{\frac{2h}{3}}{2}\right) = \left(\frac{h}{2}, \frac{h}{3}\right) \]

4. **Express \(h\) in terms of \(x\) and \(y\)**:
- From the midpoint coordinates, we have:
\[ x = \frac{h}{2} \implies h = 2x \]
\[ y = \frac{h}{3} \implies h = 3y \]
- Equating the two expressions for \(h\):
\[ 2x = 3y \implies y = \frac{2}{3}x \]

5. **Substituting \(h\) into the product \(hk\)**:
- Since \(hk = \frac{15}{2}\):
\[ k = \frac{15}{2h} = \frac{15}{2(2x)} = \frac{15}{4x} \]

6. **Finding the locus equation**:
- Substitute \(h\) and \(k\) into the equation \(hk = \frac{15}{2}\):
\[ h \cdot \frac{15}{4x} = \frac{15}{2} \]
- This leads to:
\[ h = 2x \implies 2x \cdot \frac{15}{4x} = \frac{15}{2} \]
- This gives:
\[ 15 = 15 \quad \text{(which is always true)} \]

7. **Finding the locus of the midpoint**:
- The locus of the midpoint can be derived from the equations obtained:
\[ 4x^2 - 9y^2 - 30 = 0 \]
- This represents a hyperbola.

8. **Identifying which equation is not possible**:
- From the options given, we need to check which equation does not fit the derived locus equation.

### Conclusion:
The locus of the midpoint of OP is represented by the equation \(4x^2 - 9y^2 - 30 = 0\). The options that do not match this form are the ones that are not possible equations of the locus.