Question

If the points (0,0), $(3,\sqrt{3})$ ,(p,q) from 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)

$\sqrt{3}$

Consider the given point

$A(x_1,y_1)=(0,0)$

$B(x_2,y_2)=(3,\sqrt{3})$

$C(x_3,y_3)=(p,q)$

Then,

We know that it is an equilateral triangle,

$AB=BC$

$\Rightarrow \sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}=\sqrt{{(x_2-x_3)}^2+{(y_2-y_3)}^2}$

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

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

$\text{Squaring both side and we get,}$

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

$\Rightarrow p^2+q^2-6p-2\sqrt{3}q=0$ ....... (1)

Again,

$AB=CA$

$\Rightarrow \sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}=\sqrt{{(x_3-x_1)}^2+{(y_3-y_1)}^2}$

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

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

$\text{squaring both side and we get,}$

$p^2+q^2=12$…… (2)

By equation (1) and (2) to, and we get,

$12-6p-2\sqrt{3}q=0$

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

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

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

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

Put the value of p in equation (2) and we get,

$p^2+q^2=12$

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

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

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

$\begin{align}

$\Rightarrow 12(q^2-\sqrt{3}q-6)=0$

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

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

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

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

Then, $q=2\sqrt{3},q=-\sqrt{3}$

Now, According to given question,

$q=q_1+q_2$

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

$q=\sqrt{3}$

Hence, this is the answer.