If (a, b) be an end of a diagonal of a square and the other diagonal has the equation `x-y=a`, then another vertex of the square can be

Options

A. `(a-b,a)`
B. `(a,0)`
C. `(0,-a)`
D. `(a+b,b)`

Answer

To solve the problem, we need to find another vertex of the square given that one vertex is at (a, b) and the equation of the other diagonal is \( x - y = a \).

### Step-by-Step Solution:

1. **Identify the given information:**
- One vertex of the square is \( A(a, b) \).
- The equation of the other diagonal \( BD \) is given by \( x - y = a \).

2. **Rewrite the equation of diagonal BD:**
- From the equation \( x - y = a \), we can express \( y \) in terms of \( x \):
\[ y = x - a \tag{1} \]

3. **Determine the slope of diagonal AC:**
- Since \( AC \) and \( BD \) are diagonals of the square, they are perpendicular to each other. The slope of line \( BD \) can be derived from its equation:
- The slope of line \( BD \) is \( 1 \) (since \( y = x - a \)).
- Therefore, the slope of diagonal \( AC \) is \( -1 \) (perpendicular slopes).

4. **Find the equation of diagonal AC:**
- Using the point-slope form of the equation of a line, the equation of line \( AC \) (with slope \( -1 \) and passing through point \( A(a, b) \)) is:
\[ y - b = -1(x - a) \]
- Rearranging gives:
\[ y + x - b - a = 0 \tag{2} \]

5. **Find the intersection point E of diagonals AC and BD:**
- To find point \( E \), we need to solve equations (1) and (2) simultaneously:
- Substitute \( y \) from equation (1) into equation (2):
\[ (x - a) + x - b - a = 0 \]
\[ 2x - b - 2a = 0 \implies x = \frac{b + 2a}{2} \]
- Now substitute \( x \) back into equation (1) to find \( y \):
\[ y = \frac{b + 2a}{2} - a = \frac{b}{2} + \frac{a}{2} = \frac{b + a}{2} \]
- Thus, the coordinates of point \( E \) are:
\[ E\left(\frac{b + 2a}{2}, \frac{b + a}{2}\right) \]

6. **Use the midpoint formula to find the other vertex:**
- Let the coordinates of the other vertex \( C \) be \( (x_3, y_3) \).
- The midpoint \( E \) can also be expressed in terms of \( A \) and \( C \):
\[ E = \left(\frac{a + x_3}{2}, \frac{b + y_3}{2}\right) \]
- Setting the two expressions for \( E \) equal gives us:
\[ \frac{a + x_3}{2} = \frac{b + 2a}{2} \quad \text{and} \quad \frac{b + y_3}{2} = \frac{b + a}{2} \]
- Solving these equations:
- From the first equation:
\[ a + x_3 = b + 2a \implies x_3 = b + a \]
- From the second equation:
\[ b + y_3 = b + a \implies y_3 = a \]

7. **Conclusion:**
- The coordinates of the other vertex \( C \) are \( (b + a, a) \).
- Thus, another vertex of the square can be \( (a + b, a) \).

### Final Answer:
The other vertex of the square can be \( (a + b, a) \).