Cartesian plane

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The distance between the orthocentre and circumcentre of triangle with vertices $(1, 2), (2, 1) and (\frac{3 + \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2}) is$

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Solution

$Let A (1, 2), B (2, 1), C (\frac{3 + \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2}) be the vertices of the triangle.$
$A B = \sqrt{{(2 - 1)}^{2} + {(1 - 2)}^{2}} = \sqrt{2}$
$\begin{array}{l} B C = \sqrt{{(\frac{3 + \sqrt{3}}{2} - 2)}^{2} + {(\frac{3 + \sqrt{3}}{2} - 1)}^{2}} \\ = \sqrt{{(\frac{\sqrt{3} - 1}{2})}^{2} + {(\frac{\sqrt{3} + 1}{2})}^{2}} \\ = \sqrt{\frac{8}{4}} = \sqrt{2} \\ C A = \sqrt{2} \end{array}$
$Δ A B C is equilateral.$
$∴ orthocentre and circumcentre coincide with centroid.$
$The distance between orthocentre and circumcentre is zero.$

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Cartesian plane

Remember concepts with our Masterclasses.

The distance between the orthocentre and circumcentre of triangle with vertices (1,2),(2,1) and 3+32,3+32 is

Let A(1,2),B(2,1),C3+32,3+32 be the vertices of the triangle. AB=2−12+1−22=2BC=3+32−22+3+32−12=3−122+3+122 =84=2CA=2ΔABC is equilateral. ∴ orthocentre and circumcentre coincide with centroid. The distance between orthocentre and circumcentre is zero.

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The distance between the orthocentre and circumcentre of triangle with vertices $(1, 2), (2, 1) and (\frac{3 + \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2}) is$