# Find coordinates of vertex of equilateral triangle

$ABC$ is an equilateral triangle , $AC = 2$

What is the value of $p$ and $q$ ?

HINT:

So, $C$ has to be $(2,0)$

Now, equating the squares of lengths of the sides $$(p-0)^2+(q-0)^2=(p-2)^2+(q-0)^2$$

Solve for $p$ and find $q$ from $p^2+q^2=2^2$

• Can please explain me complete answer , it will be great help ! – SSK Jun 10 '13 at 14:29
• @SSK, what was the explanation you needed? – lab bhattacharjee Jun 10 '13 at 14:34
• how i can find the $p$ and $q$ ? .. using above equation – SSK Jun 10 '13 at 14:35
• @SSK, from the first relation, $p^2=p^2-4p+4\implies p=1$ and from the second $1^2+q^2=4\implies q=\sqrt3$ as $q>0$ – lab bhattacharjee Jun 10 '13 at 14:37

By symmetry, $p=1$. Then use Pythagoras to get $q$ from $2^2=1^2+q^2$

since all angles in a equilateral triangle is 60 degrees ,

therefore slope for AB = tan60 = sqrt(3)

        slope for BC = - tan60 = -sqrt(3)


eqn of AB : y-0 = sqrt(3)(x-0)

        y-(sqrt[3])x=0   - 1


eqn of BC : y-0 = -sqrt(3) (x-2)

        y+(sqrt[3])x = 2*sqrt(3)    - 2


solving eqns 1 and 2, q = sqrt(3) , p=1