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The vertices of a triangle are $A(0,0), B(0,2)$ and $C(2,0)$, then find the distance between its orthocentre and circumcentre.
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Verified Answer
The correct answer is:
$\sqrt{2}$ units
Given $A(0,0), B(0,2) C(2,0)$,
$\overline{A C}$ is a horizontal line and
$\overline{A B}$ is a vertical line.
Given, vertices of $\triangle A B C$ are the vertices of right-angled triangle, right-angled at $A$. In a rightangled triangle, $A$ is orthocentre and mid-point of $B C$ is $D\left(\frac{2+0}{2}, \frac{0+2}{2}\right)=(1,1)$, which is the
circumcentre.
$\therefore$ Required distance $=A D=\sqrt{(1-0)^2+(1-0)^2}=\sqrt{2}$ units.
$\overline{A C}$ is a horizontal line and
$\overline{A B}$ is a vertical line.
Given, vertices of $\triangle A B C$ are the vertices of right-angled triangle, right-angled at $A$. In a rightangled triangle, $A$ is orthocentre and mid-point of $B C$ is $D\left(\frac{2+0}{2}, \frac{0+2}{2}\right)=(1,1)$, which is the
circumcentre.
$\therefore$ Required distance $=A D=\sqrt{(1-0)^2+(1-0)^2}=\sqrt{2}$ units.
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