If a point moves such that twice its distance from the axis of x exceeds its distance from the axis of y by 2, then its locus is

Options

A. `x-2y=2`
B. `x+2y=2`
C. `2y-x=2`
D. `2y-3x=5`

Answer

To find the locus of a point that moves such that twice its distance from the x-axis exceeds its distance from the y-axis by 2, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Distances**:
- The distance of a point \( P(h, k) \) from the x-axis is given by the y-coordinate, which is \( k \).
- The distance of the point \( P(h, k) \) from the y-axis is given by the x-coordinate, which is \( h \).

2. **Set Up the Given Condition**:
- According to the problem, twice the distance from the x-axis exceeds the distance from the y-axis by 2. This can be mathematically expressed as:
\[ 2k = h + 2 \]

3. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ 2k - h = 2 \]

4. **Expressing in Standard Form**:
- We can rewrite the equation in a more standard form:
\[ h - 2k = -2 \]
- To match the conventional form, we can multiply through by -1:
\[ -h + 2k = 2 \quad \text{or} \quad x - 2y = 2 \]
- Here, we replaced \( h \) with \( x \) and \( k \) with \( y \) for generalization.

5. **Conclusion**:
- The locus of the point \( P(h, k) \) is given by the equation:
\[ x - 2y = 2 \]

### Final Answer:
The locus of the point is represented by the equation:
\[ x - 2y = 2 \]