Let the fixed points be \( A(1, 2) \) and \( B(4, 5) \), which form the hypotenuse of a right-angled triangle.
Let \( P(x, y) \) be the third vertex such that triangle \( \triangle APB \) is right-angled at \( P \). Then, by the geometric property:
\[
\text{If } \angle APB = 90^\circ, \text{ then } \vec{PA} . \vec{PB} = 0
\]
Step 1: Compute vectors:
\[
\vec{PA} = (1 - x, 2 - y),
\vec{PB} = (4 - x, 5 - y)
\]
Step 2: Apply the dot product condition:
\[
\vec{PA} . \vec{PB} = 0 \Rightarrow (1 - x)(4 - x) + (2 - y)(5 - y) = 0
\]
\[
(1 - x)(4 - x) = 4 - 5x + x^2,
(2 - y)(5 - y) = 10 - 7y + y^2
\]
\[
\Rightarrow x^2 + y^2 - 5x - 7y + 14 = 0
\]
This is the required locus of the third vertex.