Find the coordinate of the third vertex of an equilateral triangle,whose two vertices are (3,4) and (-2,3) ?______________solve this in short using tricks in two to three steps if not solve long ?

1 Answer
Jan 5, 2018

The two points are (1/2+sqrt3/2,7/2-(5sqrt3)/2) and (1/2-sqrt3/2,7/2+(5sqrt3)/2)

Explanation:

The two points given are A(3,4) and B(-2,3). Let the third point be C(x,y).

Step 1

Now as AB=BC=CA, we have AB^2=BC^2=CA^2
.
or (x-3)^2+(y-4)^2=(x+2)^2+(y-3)^2

or x^2-6x+9+y^2-8y+16=x^2+4x+4+y^2-6y+9

or 10x+2y=12 or 5x+y=6 i.e. y=6-5x

Step 2

also (x-3)^2+(y-4)^2=(3+2)^2+(4-3)^2

or x^2-6x+9+y^2-8y+16=26

or x^2+y^2-6x-8y-1=0

or x^2+(6-5x)^2-6x-8(6-5x)-1=0

or x^2+36+25x^2-60x-6x-48+40x-1=0

or 26x^2-26x-13=0 or 2x^2-2x-1=0

Hence x=(2+-sqrt(4+8))/4=1/2+-sqrt3/2

and y=6-5(1/2+sqrt3/2)=7/2-(5sqrt3)/2

or y=6-5(1/2-sqrt3/2)=7/2+(5sqrt3)/2

and the two points are (1/2+sqrt3/2,7/2-(5sqrt3)/2) and (1/2-sqrt3/2,7/2+(5sqrt3)/2)

graph{((x-3)^2+(y-4)^2-0.05)((x+2)^2+(y-3)^2-0.05)((x-1/2-sqrt3/2)^2+(y-7/2+(5sqrt3)/2)^2-0.05)((x-1/2+sqrt3/2)^2+(y-7/2-(5sqrt3)/2)^2-0.05)(5x+y-6)=0 [-9.875, 10.125, -1.64, 8.36]}