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 1: Identify the points A and B
Given points are:
- \( A(3, 1) \)
- \( B(-5, 7) \)

### Step 2: Reflect point A across the line \( y = x \)
The reflection of point \( A(3, 1) \) across the line \( y = x \) will be \( A'(1, 3) \). This is because the coordinates are swapped when reflecting over the line \( y = x \).

### Step 3: Set point P on the line \( y = x \)
Let the point \( P \) on the line \( y = x \) be represented as \( P(a, a) \).

### Step 4: Write the distance expressions
The distances from \( P \) to \( A \) and \( B \) are given by:
- \( PA = \sqrt{(a - 3)^2 + (a - 1)^2} \)
- \( PB = \sqrt{(a + 5)^2 + (a - 7)^2} \)

### Step 5: Minimize the sum of distances \( PA + PB \)
Instead of minimizing \( PA + PB \) directly, we can minimize \( PA' + PB \) since \( PA + PB \) will have the same minimum point as \( PA' + PB \).

### Step 6: Write the distance expression for \( PA' \)
The distance from \( P \) to \( A' \) is:
\[ PA' = \sqrt{(a - 1)^2 + (a - 3)^2} \]

### Step 7: Set up the equation
We need to minimize:
\[ PA' + PB = \sqrt{(a - 1)^2 + (a - 3)^2} + \sqrt{(a + 5)^2 + (a - 7)^2} \]

### Step 8: Use calculus or geometric properties
To find the point \( P \) that minimizes \( PA + PB \), we can use the property that the minimum distance occurs when \( P \), \( A' \), and \( B \) are collinear.

### Step 9: Find the slope of line \( A'B \)
Using points \( A'(1, 3) \) and \( B(-5, 7) \):
- Slope \( m = \frac{7 - 3}{-5 - 1} = \frac{4}{-6} = -\frac{2}{3} \)

### Step 10: Write the equation of line \( A'B \)
Using point-slope form:
\[ y - 3 = -\frac{2}{3}(x - 1) \]
This simplifies to:
\[ y = -\frac{2}{3}x + \frac{2}{3} + 3 \]
\[ y = -\frac{2}{3}x + \frac{11}{3} \]

### Step 11: Find the intersection with line \( y = x \)
Set \( y = x \):
\[ x = -\frac{2}{3}x + \frac{11}{3} \]
\[ \frac{5}{3}x = \frac{11}{3} \]
\[ x = \frac{11}{5} \]

### Step 12: Find the coordinates of point P
Since \( P \) lies on the line \( y = x \):
\[ P = \left(\frac{11}{5}, \frac{11}{5}\right) \]

### Final Answer
Thus, the point \( P \) is:
\[ P\left(\frac{11}{5}, \frac{11}{5}\right) \]

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