Let the equations of perpendicular bisectors of sides AC and AB of `/_\ ABC` be `x+y=3 and x-y=1` respectively. Then vertex A is is (0,0).
Length of side BC of the `DeltaABC` is

Options

A. (a) `sqrt2`
B. (b) `sqrt8`
C. (c) `sqrt12`
D. (d) `sqrt20`

Answer

To find the length of side BC of triangle ABC given the equations of the perpendicular bisectors of sides AC and AB, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the equations of the perpendicular bisectors:**
- The equation of the perpendicular bisector of side AC is:
\[ x + y = 3 \]
- The equation of the perpendicular bisector of side AB is:
\[ x - y = 1 \]

2. **Find the coordinates of point B:**
- The equation of line AB can be derived from the slope of the perpendicular bisector. The slope of the perpendicular bisector \(x - y = 1\) is 1, hence the slope of line AB is \(-1\) (negative reciprocal).
- Since point A is at (0, 0), we can write the equation of line AB using point-slope form:
\[ y - 0 = -1(x - 0) \implies y = -x \implies x + y = 0 \]
- Now we have two equations to solve for point B:
\[ x + y = 0 \quad \text{(equation of AB)} \]
\[ x - y = 1 \quad \text{(equation of perpendicular bisector of AB)} \]
- Adding these two equations:
\[ (x + y) + (x - y) = 0 + 1 \implies 2x = 1 \implies x = \frac{1}{2} \]
- Substituting \(x = \frac{1}{2}\) into \(x + y = 0\):
\[ \frac{1}{2} + y = 0 \implies y = -\frac{1}{2} \]
- Thus, the coordinates of point B are:
\[ B\left(\frac{1}{2}, -\frac{1}{2}\right) \]

3. **Find the coordinates of point C:**
- The equation of line AC can also be derived from the slope of the perpendicular bisector \(x + y = 3\). The slope of this line is -1, hence the slope of line AC is 1.
- Using point-slope form again, with point A at (0, 0):
\[ y - 0 = 1(x - 0) \implies y = x \implies x - y = 0 \]
- Now we have two equations to solve for point C:
\[ x - y = 0 \quad \text{(equation of AC)} \]
\[ x + y = 3 \quad \text{(equation of perpendicular bisector of AC)} \]
- Adding these two equations:
\[ (x - y) + (x + y) = 0 + 3 \implies 2x = 3 \implies x = \frac{3}{2} \]
- Substituting \(x = \frac{3}{2}\) into \(x - y = 0\):
\[ \frac{3}{2} - y = 0 \implies y = \frac{3}{2} \]
- Thus, the coordinates of point C are:
\[ C\left(\frac{3}{2}, \frac{3}{2}\right) \]

4. **Calculate the length of side BC using the distance formula:**
- The distance \(d\) between points \(B\left(\frac{1}{2}, -\frac{1}{2}\right)\) and \(C\left(\frac{3}{2}, \frac{3}{2}\right)\) is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
- Substituting the coordinates:
\[ d = \sqrt{\left(\frac{3}{2} - \frac{1}{2}\right)^2 + \left(\frac{3}{2} - \left(-\frac{1}{2}\right)\right)^2} \]
\[ = \sqrt{\left(1\right)^2 + \left(2\right)^2} = \sqrt{1 + 4} = \sqrt{5} \]

### Final Answer:
The length of side BC of triangle ABC is:
\[ \sqrt{5} \]