Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

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

MCQ

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

#### Solution

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.

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Q 3 | Page 133