The equation of the perpendicular bisector of the side AB of a triangle ABC is `x-y+5=0`. If the point A is `(1,2)`, find the co-ordinates of the point `B`.

Answer

To find the coordinates of point B given that the equation of the perpendicular bisector of side AB of triangle ABC is \( x - y + 5 = 0 \) and point A is \( (1, 2) \), we can follow these steps:

### Step 1: Identify the midpoint D of segment AB
Since the line given is the perpendicular bisector of segment AB, it will pass through the midpoint D of segment AB. Let the coordinates of point B be \( (x_B, y_B) \). The coordinates of the midpoint D can be calculated using the midpoint formula:

\[ D = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2} \right) \]

Given \( A(1, 2) \), we have:

\[ D = \left( \frac{1 + x_B}{2}, \frac{2 + y_B}{2} \right) \]

### Step 2: Substitute D into the equation of the perpendicular bisector
Since point D lies on the line \( x - y + 5 = 0 \), we can substitute the coordinates of D into this equation:

\[ \frac{1 + x_B}{2} - \frac{2 + y_B}{2} + 5 = 0 \]

### Step 3: Simplify the equation
Multiplying through by 2 to eliminate the fraction gives:

\[ 1 + x_B - 2 - y_B + 10 = 0 \]

This simplifies to:

\[ x_B - y_B + 9 = 0 \]

Rearranging gives:

\[ x_B - y_B = -9 \quad \text{(Equation 1)} \]

### Step 4: Find the slope of the line AB
The slope of the line represented by the equation \( x - y + 5 = 0 \) can be found by rewriting it in slope-intercept form \( y = mx + b \):

\[ y = x + 5 \]

Thus, the slope of the perpendicular bisector is 1. Since AB is perpendicular to this line, the slope of line AB will be the negative reciprocal of 1, which is -1.

### Step 5: Use the slope to find the relationship between A and B
The slope of line AB can also be expressed as:

\[ \text{slope} = \frac{y_B - y_A}{x_B - x_A} \]

Substituting the coordinates of A:

\[ -1 = \frac{y_B - 2}{x_B - 1} \]

Cross-multiplying gives:

\[ y_B - 2 = - (x_B - 1) \]

Simplifying this yields:

\[ y_B = -x_B + 3 \quad \text{(Equation 2)} \]

### Step 6: Solve the system of equations
Now we have two equations:

1. \( x_B - y_B = -9 \) (Equation 1)
2. \( y_B = -x_B + 3 \) (Equation 2)

Substituting Equation 2 into Equation 1:

\[ x_B - (-x_B + 3) = -9 \]

This simplifies to:

\[ x_B + x_B - 3 = -9 \]

Combining like terms gives:

\[ 2x_B - 3 = -9 \]

Adding 3 to both sides:

\[ 2x_B = -6 \]

Dividing by 2:

\[ x_B = -3 \]

### Step 7: Find y_B using Equation 2
Now substitute \( x_B = -3 \) back into Equation 2 to find \( y_B \):

\[ y_B = -(-3) + 3 = 3 + 3 = 6 \]

### Conclusion
Thus, the coordinates of point B are:

\[ B(-3, 6) \]