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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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Verified Answer
The correct answer is:
$x^2+y^2-a x-a y=0$
Let $\triangle A B C$ be a right angled triangle with the ends of hypotenuse $A(0, a)$ and $C(a, 0)$
Let co-ordinates of point $B$ is $B(x, y)$
According to Pythagorous theorem,
$$
\begin{aligned}
& (A C)^2=(A B)^2+(B C)^2 \\
& \begin{aligned}
\Rightarrow\left(\sqrt{a^2+a^2}\right)^2=\left(\sqrt{(x-0)^2+(y-a)^2}\right)^2
\end{aligned} \\
& \qquad+\left(\sqrt{(x-a)^2+(y-0)^2}\right) \\
& \Rightarrow 2 a^2=x^2+(y-a)^2+(x-a)^2+y^2 \\
& \Rightarrow x^2+y^2-a x-a y=0
\end{aligned}
$$
Let co-ordinates of point $B$ is $B(x, y)$
According to Pythagorous theorem,
$$
\begin{aligned}
& (A C)^2=(A B)^2+(B C)^2 \\
& \begin{aligned}
\Rightarrow\left(\sqrt{a^2+a^2}\right)^2=\left(\sqrt{(x-0)^2+(y-a)^2}\right)^2
\end{aligned} \\
& \qquad+\left(\sqrt{(x-a)^2+(y-0)^2}\right) \\
& \Rightarrow 2 a^2=x^2+(y-a)^2+(x-a)^2+y^2 \\
& \Rightarrow x^2+y^2-a x-a y=0
\end{aligned}
$$
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