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

Options

A. mx = ny
B. n x= -my
C. nx = my
D. mx = -ny

Answer

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 will follow these steps:

### Step 1: Define the points and the locus point
Let the locus point be \( P(h, k) \). We have:
- Point \( A \) with coordinates \( (m+n, n-m) \)
- Point \( B \) with coordinates \( (m-n, n+m) \)

### Step 2: Set up the distance equations
Since the point \( P(h, k) \) is equidistant from points \( A \) and \( B \), we can set up the equation:
\[ PA = PB \]

Using the distance formula, we have:
\[ \sqrt{(h - (m+n))^2 + (k - (n-m))^2} = \sqrt{(h - (m-n))^2 + (k - (n+m))^2} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (h - (m+n))^2 + (k - (n-m))^2 = (h - (m-n))^2 + (k - (n+m))^2 \]

### Step 4: Expand both sides
Expanding both sides:
1. 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 gives:
\[ h^2 - 2h(m+n) + (m+n)^2 + k^2 - 2k(n-m) + (n-m)^2 \]

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 gives:
\[ h^2 - 2h(m-n) + (m-n)^2 + k^2 - 2k(n+m) + (n+m)^2 \]

### Step 5: Set the expanded forms equal to each other
Setting the left and right sides equal gives:
\[ h^2 - 2h(m+n) + (m+n)^2 + k^2 - 2k(n-m) + (n-m)^2 = h^2 - 2h(m-n) + (m-n)^2 + k^2 - 2k(n+m) + (n+m)^2 \]

### Step 6: Cancel out common terms
Cancelling \( h^2 \) and \( k^2 \) from both sides, we simplify to:
\[ -2h(m+n) + (m+n)^2 - 2k(n-m) + (n-m)^2 = -2h(m-n) + (m-n)^2 - 2k(n+m) + (n+m)^2 \]

### Step 7: Rearranging and simplifying
Rearranging terms and simplifying will lead to:
\[ 2k(m+n) = 2h(n+m) \]

### Step 8: Final relation
Dividing through by 2 gives:
\[ km = hn \]

### Conclusion
Thus, the locus of the point \( P(h, k) \) is given by the equation:
\[ y \cdot m = x \cdot n \]