If `A (3,1) and B (-5,7)` are any two given points, If P is a point one the line `y = x ` such that `PA + PB` is minimum then P is

Options

A. `(1,1)`
B. `(11//5, 11//5)`
C. `(7//13), 7//13)`
D. none of these

Answer

To find the point \( P \) on the line \( y = x \) such that the sum of distances \( PA + PB \) is minimized, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Points A and B:**
- Let \( A(3, 1) \) and \( B(-5, 7) \) be the given points.

2. **Find the Mirror Image of Point A:**
- Since point \( P \) lies on the line \( y = x \), we can reflect point \( A \) across the line \( y = x \) to find its mirror image \( A' \).
- The coordinates of the mirror image \( A' \) of point \( A(3, 1) \) are \( A'(1, 3) \).

3. **Set Up the Problem:**
- We need to minimize the distance \( PA + PB \). To do this, we can consider the straight line from \( A' \) to \( B \) since the shortest distance between two points is a straight line.

4. **Find the Equation of Line A'B:**
- The coordinates of \( A' \) are \( (1, 3) \) and the coordinates of \( B \) are \( (-5, 7) \).
- The slope \( m \) of line \( A'B \) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - 3}{-5 - 1} = \frac{4}{-6} = -\frac{2}{3} \]
- Using point-slope form \( y - y_1 = m(x - x_1) \):
\[ y - 3 = -\frac{2}{3}(x - 1) \]
- Rearranging gives:
\[ y - 3 = -\frac{2}{3}x + \frac{2}{3} \implies y = -\frac{2}{3}x + \frac{11}{3} \]

5. **Find the Intersection of Line A'B with y = x:**
- Set \( y = x \) in the equation of line \( A'B \):
\[ x = -\frac{2}{3}x + \frac{11}{3} \]
- Rearranging gives:
\[ x + \frac{2}{3}x = \frac{11}{3} \implies \frac{5}{3}x = \frac{11}{3} \implies x = \frac{11}{5} \]
- Since \( P \) lies on the line \( y = x \), we have \( y = \frac{11}{5} \).

6. **Conclusion:**
- Therefore, the coordinates of point \( P \) are:
\[ P\left(\frac{11}{5}, \frac{11}{5}\right) \]

### Final Answer:
The point \( P \) is \( \left(\frac{11}{5}, \frac{11}{5}\right) \).