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The incentre of the triangle with vertices $\mathrm{A}(1, \sqrt{3}), \mathrm{B}(0,0)$
and $\mathrm{C}(2,0)$ is
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and $\mathrm{C}(2,0)$ is
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Verified Answer
The correct answer is:
$\left(1, \frac{1}{\sqrt{3}}\right)$
Vertices of triangle, $\mathrm{A}(1, \sqrt{3}), \mathrm{B}(0,0), \mathrm{C}(2,0)$
observe the figure, $\mathrm{BD}=1, \mathrm{DC}=1, \mathrm{AD}=\sqrt{3}$
So, $A B=2 . A C=2$ (Using Pythagoras theorem) So, given triangle is equilateral triangle In equilateral triangle, Incentre is controid only. $\therefore$ Incentre $=\left(\frac{1+0+2}{3}, \frac{\sqrt{3}+0+0}{3}\right)$
$=\left(\frac{3}{3}, \frac{\sqrt{3}}{3}\right)$
$=\left(1, \frac{1}{\sqrt{3}}\right)$
observe the figure, $\mathrm{BD}=1, \mathrm{DC}=1, \mathrm{AD}=\sqrt{3}$
So, $A B=2 . A C=2$ (Using Pythagoras theorem) So, given triangle is equilateral triangle In equilateral triangle, Incentre is controid only. $\therefore$ Incentre $=\left(\frac{1+0+2}{3}, \frac{\sqrt{3}+0+0}{3}\right)$
$=\left(\frac{3}{3}, \frac{\sqrt{3}}{3}\right)$
$=\left(1, \frac{1}{\sqrt{3}}\right)$
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