A) (0, - 4)
B) (0, 4)
C) (4, 0)
D) (- 4, 0)
Correct Answer: C
Solution :
\[l=\sqrt{4+12}=4\,\,\Rightarrow \,\,{{a}^{2}}+{{b}^{2}}=16\] and \[{{(a-2)}^{2}}+{{(b-2\sqrt{3})}^{2}}=16\,\,\Rightarrow \,\,a+\sqrt{3}b=4\] Hence\[(a,\,\,b)=(4,\,\,0)\].You need to login to perform this action.
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