Question

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

Answer 1

Step-by-step explanation:

Let ABC be the equilateral triangle such that,

A=(3,0),B=(6,0) and C=(x,y)

Distance formula:

√{(x2-x1)^+(y2-y1)^2}

we know that ,

AB=BC=AC

By distance formula we get,

AB=BC=AC=3units

AC=BC

√{(3-x)^2+y^2}=√{(6-x)^2+y^2}

9+x^2-6x+y^2=36+x^2-12x+y^2

6x=27

x=27/6=9/2

BC=3units

√{(6-27/6)^2+y^2}=3

{(36-27)/6}^2+y^2=9

(9/6)^2+y^2=9

(3/2)^2+y^2=9

9/4+y^2=9

9+4y^2=36

4y^2=27

y^2=27/4

y=√(27/4)

y=3√3/2

(x,y)=(9/2,3√3/2)

Hence third vertex of equilateral triangle=C=(9/2,3√3/2)