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]}