Question

if the two vertices of an equilateral triangle are 3, 0 and 6,0 find the third vertex

Answer 1

here I have used distance formula and the property of equilateral triangle.

Answer 2

Answer:  The required co-ordinates of the third vertex are $\left(\dfrac{9}{2},\dfrac{3\sqrt3}{2}\right)$ or $\left(\dfrac{9}{2},-\dfrac{3\sqrt3}{2}\right)$.

Step-by-step explanation:  Given that the co-ordinates of the two vertices of an equilateral triangle are A(3, 0) and B(6, 0).

We are to find the co-ordinates of the third vertex.

Let C(x, y) be the co-ordinates of the third vertex.

We know that the lengths of all he sides of an equilateral triangle are equal, so we have

$BC=AC\\\\\Rightarrow \sqrt{(x-6)^2+(y-0)^2}=\sqrt{(x-3)^2+(y-0)^2}\\\\\Rightarrow x^2-12x+36+y^2=x^2-6x+9+y^2\\\\\Rightarrow 12x-6x=36-9\\\\\Rightarrow 6x=27\\\\\Rightarrow x=\dfrac{9}{2}.$

Also,

$AB=AC\\\\\Rightarrow \sqrt{(6-3)^2+(0-0)^2}=\sqrt{(\dfrac{9}{2}-3)^2+(y-0)^2}\\\\\\\Rightarrow 9=\dfrac{9}{4}+y^2\\\\\\\Rightarrow y^2=\dfrac{27}{4}\\\\\\\Rightarrow y=\pm\dfrac{3\sqrt3}{2}.$

Thus, the required co-ordinates of the third vertex are $\left(\dfrac{9}{2},\dfrac{3\sqrt3}{2}\right)$ or $\left(\dfrac{9}{2},-\dfrac{3\sqrt3}{2}\right)$.