question_answer

Find the circum-centre of the triangle whose vertices are \[(0,\text{ }0),\text{ (}3,\text{ }\sqrt{3})\] and \[(0,2\sqrt{3})\] .

A)  \[(1,\sqrt{3})\]

B)  \[(\sqrt{3},\sqrt{3})\]

C)  \[(2\sqrt{3},1)\]            

D)  \[(2,\sqrt{3})\]

Correct Answer: A

Solution :

(a): Let \[O(0,0),A(3,\sqrt{3}),B(0,2\sqrt{3})\] \[OB=\sqrt{{{(2\sqrt{3})}^{2}}}=\sqrt{12}\]; \[AO=\sqrt{{{3}^{2}}+{{(\sqrt{3})}^{2}}}=\sqrt{12}\] \[AB=\sqrt{{{3}^{2}}+{{(\sqrt{3}-2\sqrt{3})}^{2}}}=\sqrt{12}\]; \[AB=\sqrt{{{3}^{2}}+{{(\sqrt{3}-2\sqrt{3})}^{2}}}=\sqrt{12}\]; \[\therefore \]\[\Delta \,ABC\] is an equilateral triangle \[\Rightarrow \]\[\left( \frac{0+3+0}{3},\frac{0+\sqrt{3}+2\sqrt{3}}{3} \right)=(1,\sqrt{3})\] is the centroid as well as circum ? centre