Find the locus of the third vertex of a right angled triangle , the ends of whose hypotenuse are (4,0) and (0,4)

Answer

To find the locus of the third vertex of a right-angled triangle whose hypotenuse endpoints are (4,0) and (0,4), we can follow these steps:

### Step 1: Identify the Points
Let the points A and B be the endpoints of the hypotenuse:
- A = (4, 0)
- B = (0, 4)

Let the third vertex, C, be represented by the coordinates (x, y).

### Step 2: Apply the Pythagorean Theorem
According to the Pythagorean theorem for a right-angled triangle, we have:
\[ AC^2 + BC^2 = AB^2 \]

### Step 3: Calculate the Lengths
Now, we need to calculate the lengths of the sides:
1. Length of AB (hypotenuse):
\[ AB = \sqrt{(4 - 0)^2 + (0 - 4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \]
Therefore,
\[ AB^2 = (4\sqrt{2})^2 = 32 \]

2. Length of AC:
\[ AC = \sqrt{(x - 4)^2 + (y - 0)^2} = \sqrt{(x - 4)^2 + y^2} \]
Thus,
\[ AC^2 = (x - 4)^2 + y^2 \]

3. Length of BC:
\[ BC = \sqrt{(x - 0)^2 + (y - 4)^2} = \sqrt{x^2 + (y - 4)^2} \]
Thus,
\[ BC^2 = x^2 + (y - 4)^2 \]

### Step 4: Set Up the Equation
Now substituting these lengths into the Pythagorean theorem:
\[ AC^2 + BC^2 = AB^2 \]
This gives us:
\[ (x - 4)^2 + y^2 + x^2 + (y - 4)^2 = 32 \]

### Step 5: Expand and Simplify
Now we will expand the equation:
1. Expand \( (x - 4)^2 \):
\[ (x - 4)^2 = x^2 - 8x + 16 \]
2. Expand \( (y - 4)^2 \):
\[ (y - 4)^2 = y^2 - 8y + 16 \]
3. Substitute back into the equation:
\[ (x^2 - 8x + 16) + y^2 + (x^2 + y^2 - 8y + 16) = 32 \]
Combine like terms:
\[ 2x^2 + 2y^2 - 8x - 8y + 32 = 32 \]
Simplifying gives:
\[ 2x^2 + 2y^2 - 8x - 8y = 0 \]
Dividing the entire equation by 2:
\[ x^2 + y^2 - 4x - 4y = 0 \]

### Step 6: Final Equation
Rearranging the equation gives us the locus of the third vertex:
\[ x^2 + y^2 - 4x - 4y = 0 \]

### Conclusion
The locus of the third vertex of the right-angled triangle is given by the equation:
\[ x^2 + y^2 - 4x - 4y = 0 \]