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(A) Let the third vertex be $P(x, y)$.Since the triangle is a right-angled triangle with the hypotenuse connecting $(2, 0)$ and $(0, 2)$,the angle at

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$x^2 + y^2 - 2x - 2y = 0$

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(A) Let the third vertex be $P(x, y)$.Since the triangle is a right-angled triangle with the hypotenuse connecting $(2, 0)$ and $(0, 2)$,the angle at vertex $P$ is $90^{\circ}$.The product of the slopes of the two sides meeting at $P$ must be $-1$.Slope $m_1 = \frac{y - 0}{x - 2} = \frac{y}{x - 2}$.Slope $m_2 = \frac{y - 2}{x - 0} = \frac{y - 2}{x}$.Setting $m_1 	imes m_2 = -1$:$\frac{y}{x - 2} 	imes \frac{y - 2}{x} = -1$$y(y - 2) = -x(x - 2)$$y^2 - 2y = -x^2 + 2x$$x^2 + y^2 - 2x - 2y = 0$.

If the endpoints of the hypotenuse of a right-angled triangle are $(2, 0)$ and $(0, 2)$,find the locus of its third vertex.

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