First slide

Cartesian plane

Question

The incenter of the triangle with vertices $(1, \sqrt{3})$ , (0, 0), and (2, 0) is

Easy

A

$(1, \frac{\sqrt{3}}{2})$
B

$(\frac{2}{3}, \frac{1}{\sqrt{3}})$
C

$(\frac{2}{3}, \frac{\sqrt{3}}{2})$
D

$(1, \frac{1}{\sqrt{3}})$

Solution

Here, AB = BC = CA = 2. So, it is an equilateral triangle and the incenter coincides with the centroid.
Therefore, the centroid is $(\frac{0 + 1 + 2}{3}, \frac{0 + 0 + \sqrt{3}}{3}) \equiv (1, \frac{1}{\sqrt{3}})$

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