Question:
The incentre of the triangle with vertices $(1, \sqrt{3}), (0,0)$ and $ (2,0) is $
The incentre of the triangle with vertices $(1, \sqrt{3}), (0,0)$ and $ (2,0) is $
Updated On: Jun 14, 2022
- $\Big(1, \frac{\sqrt{3}}{2}\Big)$
- $\Big(\frac{2}{3} \frac{1}{\sqrt{3}}\Big)$
- $\Big(\frac{2}{3} \frac{\sqrt{3}}{2}\Big)$
- $\Big(1,\frac{1}{\sqrt{3}}\Big)$
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The Correct Option is D
Solution and Explanation
Let the vertices of triangle be $A(1, \sqrt{3}), 5(0.0)$ and $C(2,0)$. Here, $AB = BC = CA=2$
Therefore, it is an equilateral triangle. So, the in centre coincides with centroid.
$\therefore \hspace25mm I\equiv \Bigg(\frac{0 + 1 + 2}{3},\frac{0+0+\sqrt{3}}{3}\Bigg)$
$\Rightarrow \hspace25mm I \equiv(1/\sqrt{3})$
Therefore, it is an equilateral triangle. So, the in centre coincides with centroid.
$\therefore \hspace25mm I\equiv \Bigg(\frac{0 + 1 + 2}{3},\frac{0+0+\sqrt{3}}{3}\Bigg)$
$\Rightarrow \hspace25mm I \equiv(1/\sqrt{3})$
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