Question

If the points $(0,0),(3,\sqrt{3}),(p,q)$ form an equilateral triangle and $q_1,q_2$ are the two values of $q$ then $q_1+q_2=?$

• A. $2\sqrt{3}$
• B. $\sqrt{3}$
• C. $-\sqrt{3}$
• D. $0$

Answer 1

Answer: (B)

$\sqrt{3}$

Points $(0,0),(3,\sqrt{3}),(p,q)$ of an equilateral $△.$

Distance between $(0,0)$ and $(p,q)=$ Distance between $(0,0)$ and $(3,\sqrt{3})=$ Distance between $(p,q)$ and $(3,\sqrt{3}).$

$\Rightarrow \sqrt{p^2+q^2}=\sqrt{3^2+3}=\sqrt{{(p-3)}^2+{(q-\sqrt{3})}^2}$

$I$ $II$ $III$

solving $I\&II$

$\Rightarrow p^2+q^2=12⟶\left(1\right)$

solving $II\&III$

$\Rightarrow p^2+9-6p+q^2+3-2\sqrt{3}q=12$

$\Rightarrow p^2+q^2-6p-2\sqrt{3}q+12=12$

$\Rightarrow 12-6p-2\sqrt{3}q+12=12$

$\Rightarrow -6p-2\sqrt{3}q=-12$

$\Rightarrow p=\frac{6-\sqrt{3}q}{3}$

putting in $\left(I\right)$

$\Rightarrow {\left(\frac{6-\sqrt{3}q}{3}\right)}^2+q^2=12$

$\Rightarrow \frac{36+3q^2-12\sqrt{3}q}{9}+q^2=12$

$\Rightarrow 36+3q^2-12\sqrt{3}q+9q^2=108$

$\Rightarrow 12q^2-12\sqrt{3}q-72=0$

$\Rightarrow q^2-\sqrt{3}q-6=0$

$\Rightarrow q^2-2\sqrt{3}q+\sqrt{3}q-6=0$

$\Rightarrow (q-2\sqrt{3})(q+\sqrt{3})=0$

$q=2\sqrt{3}$ or $-\sqrt{3}$

If $q_1=2\sqrt{3},p=\frac{6-6}{3}=0$

$q_2=-\sqrt{3},p=\frac{6+3}{3}=3$

third vertex $(0,2\sqrt{3})$ or $(3,-\sqrt{3})$

so the value of $q=q_1+q_2$

$=2\sqrt{3}-\sqrt{3}$

$=\sqrt{3}$

Hence, the answer is $\sqrt{3}.$