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Question
The distance between the orthocentre and circumcentre of the triangle with vertices (1, 2), (2, 1) and \[\left( \frac{3 + \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2} \right)\] is
Options
0
\[\sqrt{2}\]
\[3 + \sqrt{3}\]
none of these
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Solution
0
Let A(1, 2), B(2, 1) and C \[\left( \frac{3 + \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2} \right)\] be the given points.
\[\therefore \text { AB } = \sqrt{\left( 2 - 1 \right)^2 + \left( 1 - 2 \right)^2}\]
\[ = \sqrt{2}\]
\[\text { BC } = \sqrt{\left( \frac{3 + \sqrt{3}}{2} - 2 \right)^2 + \left( \frac{3 + \sqrt{3}}{2} - 1 \right)^2}\]
\[ = \sqrt{2}\]
\[\text { AC }= \sqrt{\left( \frac{3 + \sqrt{3}}{2} - 1 \right)^2 + \left( \frac{3 + \sqrt{3}}{2} - 2 \right)^2}\]
\[ = \sqrt{2}\]
Thus, ABC is an equilateral triangle.
We know that the orthocentre and the circumcentre of an equilateral triangle are same.
So, the distance between the the orthocentre and the circumcentre of the triangle
with vertices (1, 2), (2, 1) and \[\left( \frac{3 + \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2} \right)\] is 0.
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