The ends of the hypotenuse of a right angled triangle are (6, 0) and (0, 6). Find the locus of the thrid vertex.

Answer

To find the locus of the third vertex of a right-angled triangle with the ends of the hypotenuse at points \( A(6, 0) \) and \( B(0, 6) \), we can follow these steps:

### Step 1: Define the points
Let the coordinates of the third vertex \( C \) be \( (H, K) \). The points \( A \) and \( B \) are given as \( A(6, 0) \) and \( B(0, 6) \).

### Step 2: Use the slope condition
Since \( C \) is the right angle vertex, the lines \( AC \) and \( BC \) must be perpendicular. The product of the slopes of two perpendicular lines is -1.

- The slope of line \( AC \) is given by:
\[ \text{slope of } AC = \frac{K - 0}{H - 6} = \frac{K}{H - 6} \]

- The slope of line \( BC \) is given by:
\[ \text{slope of } BC = \frac{K - 6}{H - 0} = \frac{K - 6}{H} \]

Setting the product of these slopes to -1:
\[ \frac{K}{H - 6} \cdot \frac{K - 6}{H} = -1 \]

### Step 3: Simplify the equation
Cross-multiplying gives:
\[ K(K - 6) = -H(H - 6) \]
This simplifies to:
\[ K^2 - 6K + H^2 - 6H = 0 \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ H^2 + K^2 - 6H - 6K = 0 \]

### Step 5: Completing the square
To convert this into a standard form, we complete the square for both \( H \) and \( K \):
\[ (H^2 - 6H) + (K^2 - 6K) = 0 \]

Completing the square:
\[ (H - 3)^2 - 9 + (K - 3)^2 - 9 = 0 \]
\[ (H - 3)^2 + (K - 3)^2 = 18 \]

### Step 6: Final equation of the locus
This represents a circle centered at \( (3, 3) \) with a radius of \( \sqrt{18} \). Thus, the locus of the third vertex \( C \) is given by the equation:
\[ (H - 3)^2 + (K - 3)^2 = 18 \]

### Conclusion
The locus of the third vertex \( C \) of the right-angled triangle with hypotenuse endpoints at \( (6, 0) \) and \( (0, 6) \) is a circle with the equation:
\[ x^2 + y^2 - 6x - 6y = 0 \]