Question

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 ?

Answer 1

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]}