Question

If the two vertices of an equilateral triangle be (0, 0) and (3,$\sqrt{3}$ ), then the co-ordinates of third vertex are

• A. (0, 2$\sqrt{3}$) or (3, - $\sqrt{3}$)
• B. (0, $\sqrt{3}$ ), or (3, $\sqrt{3}$ ),
• C. (0, 2) (0, 3)
• D. none of these

Answer 1

The correct option is A (0, 2$\sqrt{3}$) or (3, - $\sqrt{3}$)

$LetthepointsbeP=(p,q),Q=(x_1,y_1)=(0,0)andR=(x_2,y_2)=(3,\sqrt{3}).\because Theyareverticesofanequilateral\Delta ,\therefore PQ=QR=PRSoPQ=\sqrt{{(x_1-p)}^2+{(y_1-q)}^2}=\sqrt{{(0-p)}^2+{(0-q)}^2}=\sqrt{p^2+q^2},QR=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(3-0)}^2+{(\sqrt{3}-0)}^2}=2\sqrt{3}andPR=\sqrt{{(x_2-p)}^2+{(y_2-q)}^2}=\sqrt{{(3-p)}^2+{(\sqrt{3}-q)}^2}=\sqrt{p^{2+}+q^2-6p-2\sqrt{3}q+12}\therefore PQ=QR⟹\sqrt{p^2+q^2}=2\sqrt{3}⟹p^2+q^2=\mathrm{12........}\left(i\right)AgainQR=PR⟹\sqrt{{(3-p)}^2+{(\sqrt{3}-q)}^2}=2\sqrt{3}⟹p^2+q^2-6p-2\sqrt{3}q+12=12p^2+q^2-6p-2\sqrt{3}q=\mathrm{0............}\left(ii\right)Substituting{p}^2+q^2=12from\left(i\right)weget3p+\sqrt{3}q=6⟹q=\frac{6-3p}{\sqrt{3}}........\left(iii\right)Puttingthisvalueofqin\left(i\right)wehave{p}^2+{\left(\frac{6-3p}{\sqrt{3}}\right)}^2=12⟹12p^2-36p=0⟹p=(0,3)Puttingthesevaluesofpin\left(iii\right)wegetq=(2\sqrt{3},-\sqrt{3})\therefore P=(0,2\sqrt{3}),(3,-\sqrt{3})AnsOptionA$