मराठी

The Distance Between the Orthocentre and Circumcentre of the Triangle with Vertices (1, 2), (2, 1) and

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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

MCQ
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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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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 23: The straight lines - Exercise 23.21 [पृष्ठ १३३]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 23 The straight lines
Exercise 23.21 | Q 3 | पृष्ठ १३३

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