Suppose A, B are two points on `2x-y+3=0 and P(1,2)` is such that PA=PB. Then the mid point of AB is

Options

A. `((-1)/(5), (13)/(5))`
B. `((-7)/(5), (9)/(5))`
C. `((7)/(5), (-9)/(5))`
D. `((-7)/(5), (-9)/(5))`

Explanation

To solve the problem, we need to find the midpoint of points A and B on the line \(2x - y + 3 = 0\) such that the distances from point P(1, 2) to points A and B are equal (i.e., \(PA = PB\)).

### Step-by-Step Solution:

1. **Identify the line equation and point P**:
The line equation is given as \(2x - y + 3 = 0\). We can rewrite this in slope-intercept form:
\[ y = 2x + 3 \]
The point P is given as \(P(1, 2)\).

2. **Use the distance formula**:
Since \(PA = PB\), point P lies on the perpendicular bisector of segment AB. Thus, we need to find the foot of the perpendicular from point P to the line \(2x - y + 3 = 0\).

3. **Find the coefficients for the line**:
The line \(2x - y + 3 = 0\) has coefficients:
- \(a = 2\)
- \(b = -1\)
- \(c = 3\)

4. **Use the formula for the foot of the perpendicular**:
The coordinates of the foot of the perpendicular from point \(P(x_1, y_1)\) to the line \(ax + by + c = 0\) can be calculated using:
\[ \left( x, y \right) = \left( x_1 - \frac{a(ax_1 + by_1 + c)}{a^2 + b^2}, y_1 - \frac{b(ax_1 + by_1 + c)}{a^2 + b^2} \right) \]
Substituting \(P(1, 2)\) into the formula:
\[ x = 1 - \frac{2(2 \cdot 1 - 1 \cdot 2 + 3)}{2^2 + (-1)^2} \]
\[ y = 2 - \frac{-1(2 \cdot 1 - 1 \cdot 2 + 3)}{2^2 + (-1)^2} \]

5. **Calculate \(ax_1 + by_1 + c\)**:
\[ ax_1 + by_1 + c = 2 \cdot 1 - 1 \cdot 2 + 3 = 2 - 2 + 3 = 3 \]

6. **Substituting into the foot of the perpendicular formula**:
\[ x = 1 - \frac{2 \cdot 3}{5} = 1 - \frac{6}{5} = 1 - 1.2 = -0.2 = -\frac{1}{5} \]
\[ y = 2 - \frac{-1 \cdot 3}{5} = 2 + \frac{3}{5} = 2 + 0.6 = 2.6 = \frac{13}{5} \]

7. **Coordinates of the foot of the perpendicular**:
The foot of the perpendicular \(P'\) is at:
\[ P' \left(-\frac{1}{5}, \frac{13}{5}\right) \]

8. **Midpoint of AB**:
Since \(P'\) is the midpoint of segment AB, the coordinates of the midpoint of AB are:
\[ \text{Midpoint} = P' \left(-\frac{1}{5}, \frac{13}{5}\right) \]

### Final Answer:
The midpoint of points A and B is:
\[ \left(-\frac{1}{5}, \frac{13}{5}\right) \]