Question

Find the third vertex of an equilateral triangle whose two vertices are $(2,4)$ and $(2,6)$.

• A. $(2+\sqrt{3},5)$ or $(2-\sqrt{3},5)$
• B. $(2+\sqrt{3},5)$ or $(2-\sqrt{7},5)$
• C. $(2+\sqrt{7},5)$ or $(2-\sqrt{3},5)$
• D. None of these

Answer 1

The correct option is A $(2+\sqrt{3},5)$ or $(2-\sqrt{3},5)$

Length of side of equilateral triangle is $AB=BC=CA=2$.

Let's take the third vertex to be $C(x,y)$

Then $AC=BC=>A{C}^2=B{C}^2$.

By distance formula:

$\sqrt{(}(x-2{)}^2+(y-4{)}^2)=\sqrt{(}(x-2{)}^2+(y-6{)}^2)$

$=>-4y+4=-12y+36$

$=>y=5$

Now $A{C}^2=4=\left(\right(x-2{)}^2+(y-4{)}^2)$.

Putiing $y=5$:

$=>x^2-4x+1=0$.

So, solving the quadratic we get:

$x=2+\sqrt{3}$ or $2-\sqrt{3}$.

So, required point is $(2+\sqrt{3},5)$ and $(2-\sqrt{3},5)$.