The locus represented by the equation `(x-y+c)^(2)+(x+y-c)^(2)=0` is

Options

A. a line parallel to x-axis
B. a point
C. a pair of straight lines
D. a line parallel of y - axis

Answer

To solve the equation \((x - y + c)^2 + (x + y - c)^2 = 0\) and find the locus it represents, we can follow these steps:

### Step 1: Analyze the equation
The given equation is \((x - y + c)^2 + (x + y - c)^2 = 0\).

### Step 2: Understand the implications of the equation
Since both terms in the equation are squares, they are always non-negative. The only way their sum can equal zero is if both terms are individually equal to zero.

### Step 3: Set each term to zero
We set each squared term to zero:
1. \( (x - y + c)^2 = 0 \)
2. \( (x + y - c)^2 = 0 \)

### Step 4: Solve the first equation
From the first equation:
\[ x - y + c = 0 \implies x - y = -c \implies x = y - c \]

### Step 5: Solve the second equation
From the second equation:
\[ x + y - c = 0 \implies x + y = c \implies x = c - y \]

### Step 6: Set the two expressions for \(x\) equal to each other
Now, we have two expressions for \(x\):
1. \(x = y - c\)
2. \(x = c - y\)

Setting them equal gives:
\[ y - c = c - y \]

### Step 7: Solve for \(y\)
Rearranging the equation:
\[ y + y = c + c \implies 2y = 2c \implies y = c \]

### Step 8: Substitute back to find \(x\)
Now substitute \(y = c\) back into either expression for \(x\):
Using \(x = y - c\):
\[ x = c - c = 0 \]

### Conclusion
Thus, we have found that the locus represented by the equation is the point \((0, c)\).

### Final Answer
The locus represented by the equation \((x - y + c)^2 + (x + y - c)^2 = 0\) is the point \((0, c)\).

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