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Question

(A) Let the third vertex be $B(x, y)$. The ends of the hypotenuse are $A(0, a)$ and $C(a, 0)$.Since $	riangle ABC$ is a right-angled triangle at $B$,

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the third vertex be $B(x, y)$. The ends of the hypotenuse are $A(0, a)$ and $C(a, 0)$.Since $	riangle ABC$ is a right-angled triangle at $B$,the angle $\angle ABC = 90^{\circ}$.The slope of $AB$ is $m_1 = \frac{y-a}{x-0} = \frac{y-a}{x}$.The slope of $BC$ is $m_2 = \frac{y-0}{x-a} = \frac{y}{x-a}$.Since $AB \perp BC$,the product of their slopes is $-1$:$\left(\frac{y-a}{x}ight) 	imes \left(\frac{y}{x-a}ight) = -1$$y(y-a) = -x(x-a)$$y^2 - ay = -x^2 + ax$$x^2 + y^2 - ax - ay = 0$.

If the ends of the hypotenuse of a right-angled triangle are $(0, a)$ and $(a, 0)$,then the locus of the third vertex is:

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