What is the locus of a point which is equidistant from the point `(m+n, n-m)` and the point `(m-n, n+m)` ?

Options

A. `mx=ny`
B. `nx=-my`
C. `nx=my`
D. `mx=-ny`

Explanation

To find the locus of a point that is equidistant from the points \( A(m+n, n-m) \) and \( B(m-n, n+m) \), we can follow these steps:

### Step 1: Define the point
Let the point whose locus we want to find be \( P(h, k) \).

### Step 2: Use the distance formula
We need to set the distances from point \( P \) to points \( A \) and \( B \) equal to each other:
\[ PA = PB \]
Using the distance formula, we can express these distances:
\[ PA = \sqrt{(h - (m+n))^2 + (k - (n-m))^2} \]
\[ PB = \sqrt{(h - (m-n))^2 + (k - (n+m))^2} \]

### Step 3: Set the distances equal
Now, we set the two distances equal to each other:
\[ \sqrt{(h - (m+n))^2 + (k - (n-m))^2} = \sqrt{(h - (m-n))^2 + (k - (n+m))^2} \]

### Step 4: Square both sides
To eliminate the square roots, we square both sides:
\[ (h - (m+n))^2 + (k - (n-m))^2 = (h - (m-n))^2 + (k - (n+m))^2 \]

### Step 5: Expand both sides
Now we expand both sides:
Left side:
\[ (h - (m+n))^2 = h^2 - 2h(m+n) + (m+n)^2 \]
\[ (k - (n-m))^2 = k^2 - 2k(n-m) + (n-m)^2 \]
Combining these gives:
\[ h^2 - 2h(m+n) + (m+n)^2 + k^2 - 2k(n-m) + (n-m)^2 \]

Right side:
\[ (h - (m-n))^2 = h^2 - 2h(m-n) + (m-n)^2 \]
\[ (k - (n+m))^2 = k^2 - 2k(n+m) + (n+m)^2 \]
Combining these gives:
\[ h^2 - 2h(m-n) + (m-n)^2 + k^2 - 2k(n+m) + (n+m)^2 \]

### Step 6: Simplify the equation
Now we can simplify both sides:
After canceling \( h^2 \) and \( k^2 \) from both sides, we have:
\[ -2h(m+n) - 2k(n-m) + (m+n)^2 + (n-m)^2 = -2h(m-n) - 2k(n+m) + (m-n)^2 + (n+m)^2 \]

### Step 7: Rearrange and combine like terms
Rearranging and combining like terms leads to:
\[ 2h(m-n) - 2h(m+n) + 2k(n+m) - 2k(n-m) = 0 \]
This simplifies to:
\[ 2h(-2n) + 2k(2m) = 0 \]
Thus, we can factor out the 2:
\[ -hn + km = 0 \]
or
\[ mh = kn \]

### Conclusion
The locus of the point \( P(h, k) \) is given by the equation:
\[ nx = my \]
where \( (h, k) \) corresponds to \( (x, y) \).

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