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

Answer: (A)

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

Given that, Two vertices of an equilateral triangle are $(0,0)$ and $(3,\sqrt{3})$

Let the third vertex of the equilateral triangle be $(x,y)$

Distance between $(0,0)$ and$(x,y)$ =Distance between$(0,0)$ and$(3,\sqrt{3})$ =Distance between$(x,y)$and $(3,\sqrt{3})$.

$\sqrt{x^2+y^2}=\sqrt{{(x-3)}^2+{(y-\sqrt{3})}^2}$

$x^2+y^2={(x-3)}^2+{(y-\sqrt{3})}^2$

$x^2+y^2=12$

$x^2+9-6x+y^2+3-2\sqrt{3}y=12$

$24-6x-2\sqrt{3}y=12$

$-6x-2\sqrt{3}y=-12$

$6x+2\sqrt{3}y=12$

$3x+\sqrt{3}y=6$

$x=\frac{(6-\sqrt{3})}{3}$

$\Rightarrow {\left[\frac{6-\sqrt{3}y}{3}\right]}^3+y^2=12$

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

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

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

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

$\Rightarrow y=2\sqrt{3}or-\sqrt{3}$

$ify=2\sqrt{3},x=\frac{(6-6)}{3}=0$

$ify=-\sqrt{3},x=\frac{(6+3)}{3}=3$

So, the third vertex of the equilateral triangle $=(0,2\surd 3)or(3,-\surd 3)$.