The vertices of a triangle are A(0,0), B(0,2) | vedclass

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(B) Given vertices are $A(0,0)$,$B(0,2)$,and $C(2,0)$.Since $\overline{AC}$ lies on the $x$-axis and $\overline{AB}$ lies on the $y$-axis,the triangle

Vedclass · Accepted Answer

$\sqrt{2} 	ext{ units}$

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(B) Given vertices are $A(0,0)$,$B(0,2)$,and $C(2,0)$.Since $\overline{AC}$ lies on the $x$-axis and $\overline{AB}$ lies on the $y$-axis,the triangle is a right-angled triangle with the right angle at vertex $A(0,0)$.In a right-angled triangle,the orthocentre is the vertex where the right angle is formed,so the orthocentre is $H(0,0)$.The circumcentre of a right-angled triangle is the midpoint of the hypotenuse $\overline{BC}$.Circumcentre $O = \left(\frac{0+2}{2}, \frac{2+0}{2}ight) = (1,1)$.The distance between the orthocentre $(0,0)$ and the circumcentre $(1,1)$ is $\sqrt{(1-0)^2 + (1-0)^2} = \sqrt{1^2 + 1^2} = \sqrt{2} 	ext{ units}$.

The vertices of a triangle are $A(0,0), B(0,2)$ and $C(2,0)$. Find the distance between its orthocentre and circumcentre.

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