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प्रश्न
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
पर्याय
0
\[\sqrt{2}\]
\[3 + \sqrt{3}\]
none of these
उत्तर
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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