Question

If the points $\left(0,0\right),\left(3,\sqrt{3}\right),\left(p,q\right)$ form an equilateral triangle $q_1,q_2$ 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

The correct option is B $\sqrt{3}$

REF.Image.

Given ABC is a equilateral

triangle

so, $A{B}^2=A{C}^2=B{C}^2$

$A{B}^2=(3-0{)}^2+(\sqrt{3}-0{)}^2=9+3=12$

$A{C}^2=(p-0{)}^2+(q-0{)}^2=p^2+q^2$

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

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

Now do$\to A{C}^2=B{C}^2$

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

$\Rightarrow 6p+2\sqrt{3}q=12$ __ (i)

Now do $\to A{B}^2=A{C}^2\Rightarrow 12=p^2+q^2$ __ (ii)

From equation (i) $\Rightarrow 6p=12-2\sqrt{3}q$

$p=\frac{12}{6}-\frac{2\sqrt{3}}{6}q=(2-\frac{2}{\sqrt{3}})$

put the value of 'p' in equation (ii)

$12=(2-\frac{q}{\sqrt{3}}{)}^2+q^2=4+\frac{q^2}{3}-\frac{4q}{\sqrt{3}}+q^2$

$12=\frac{12+q^2-4\sqrt{3}q+3q^2}{3}\Rightarrow 12\times 3=12+4q^2-4\sqrt{3}q$

$\Rightarrow 4q^2-4\sqrt{3}q=36-12=24\Rightarrow q^2-\sqrt{3}q=6$

$q^2-\sqrt{3}q-6=0$ $\left(q_1{q}^{{}_2}\right)\to$ This equation will have two roots.

so, sum of roots $=q_1+q_2=\frac{\left(\sqrt{3}\right)}{1}$

$q_1+q_2=\sqrt{3}$