A point P is equidistant from A (3, 1) and B (5, 3) and its abscissa is twice its ordinate, then its co-ordinates are.

Options

A. (2, 1)
B. (1, 2)
C. (4, 2)
D. (2,4)

Answer

To solve the problem, we need to find the coordinates of point P, which is equidistant from points A (3, 1) and B (5, 3), and has the property that its abscissa (x-coordinate) is twice its ordinate (y-coordinate).

### Step-by-Step Solution:

1. **Define the Coordinates of Point P**:
Let the coordinates of point P be (x1, y1). According to the problem, we know that:
\[ x1 = 2y1 \]

2. **Use the Distance Formula**:
Since point P is equidistant from points A and B, we can set the distances PA and PB equal to each other:
\[ PA = PB \]
Using the distance formula, we have:
\[ \sqrt{(x1 - 3)^2 + (y1 - 1)^2} = \sqrt{(x1 - 5)^2 + (y1 - 3)^2} \]

3. **Square Both Sides**:
To eliminate the square roots, we square both sides:
\[ (x1 - 3)^2 + (y1 - 1)^2 = (x1 - 5)^2 + (y1 - 3)^2 \]

4. **Expand Both Sides**:
Expanding both sides gives:
\[ (x1^2 - 6x1 + 9) + (y1^2 - 2y1 + 1) = (x1^2 - 10x1 + 25) + (y1^2 - 6y1 + 9) \]

5. **Simplify the Equation**:
Cancel \(x1^2\) and \(y1^2\) from both sides:
\[ -6x1 + 10 = -10x1 + 25 - 6y1 + 9 + 2y1 \]
Rearranging gives:
\[ 4x1 - 4y1 = 15 \]

6. **Substitute x1**:
Substitute \(x1 = 2y1\) into the equation:
\[ 4(2y1) - 4y1 = 15 \]
This simplifies to:
\[ 8y1 - 4y1 = 15 \implies 4y1 = 15 \implies y1 = \frac{15}{4} \]

7. **Find x1**:
Now substitute \(y1\) back to find \(x1\):
\[ x1 = 2y1 = 2 \times \frac{15}{4} = \frac{30}{4} = \frac{15}{2} \]

8. **Final Coordinates**:
Thus, the coordinates of point P are:
\[ P\left(\frac{15}{2}, \frac{15}{4}\right) \]

### Final Answer:
The coordinates of point P are \(\left(\frac{15}{2}, \frac{15}{4}\right)\).