A and B are the points (2, 0) and (0, 2) respectively. The coordinates of the point P on the line 2x+3y+1=0 are

Options

A. `(-7,-5)` if `|PA- PB|` is maximum
B. `((1)/(5) , (1)/(5))` if `|PA- PB|` is maximum
C. `(7,-5)` if `|PA - PB|` is minimum
D. `((1)/(5) , (1)/(5))` if `|PA- PB|` is minimum

Answer

To solve the problem, we need to find the coordinates of point P on the line \(2x + 3y + 1 = 0\) such that the difference \(PA - PB\) is maximized, where A and B are the points (2, 0) and (0, 2) respectively.

### Step 1: Find the equation of line AB
The points A(2, 0) and B(0, 2) can be used to find the equation of the line connecting them.

Using the two-point form of the line equation:
\[ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \]
Substituting \(A(2, 0)\) as \((x_1, y_1)\) and \(B(0, 2)\) as \((x_2, y_2)\):
\[ y - 0 = \frac{2 - 0}{0 - 2}(x - 2) \]
This simplifies to:
\[ y = -x + 2 \]
or rearranging gives:
\[ x + y - 2 = 0 \quad \text{(Equation of line AB)} \]

### Step 2: Write the equation of line L2
The equation of line L2 is given as:
\[ 2x + 3y + 1 = 0 \]

### Step 3: Solve the system of equations
To find the intersection of lines L1 and L2, we can solve the equations:
1. \(x + y - 2 = 0\)
2. \(2x + 3y + 1 = 0\)

From the first equation, we can express \(y\) in terms of \(x\):
\[ y = 2 - x \]

Substituting this into the second equation:
\[ 2x + 3(2 - x) + 1 = 0 \]
Expanding this gives:
\[ 2x + 6 - 3x + 1 = 0 \]
Combining like terms results in:
\[ -x + 7 = 0 \implies x = 7 \]

### Step 4: Find y-coordinate
Now substituting \(x = 7\) back into the equation for \(y\):
\[ y = 2 - 7 = -5 \]

### Step 5: Conclusion
Thus, the coordinates of point P are:
\[ P(7, -5) \]