Question

2 vertices of an equilateral triangle are(0,0) and (3,root3).Find the 3rd vertex. with steps plzz

Answer 1

step1:-let unknown point of equilatral triangle is (x, y)

step2:-we know equilatral triangle all side length same so, use this concept

step3:- above concept used by distance formula
e.g. if two point (a, b) and (r, s) given distance between =root {(a-r)^2+(b-s)^2}

if A (0,0) B (3, root3) C (x, y) points of triangle .
then AB=BC=CA
also
AB^2=BC^2=CA^2

step 4:-(0-3)^2+(0-root3)^2=(3-x)^2+(root3-y)^2=(x-0)^2+(y-0)^2

step5:-x^2+y^2=9 + 3=12
x^2 +y^2 =12 --------(1)

(3-x)^2+(root3-y)^2=12
9+x^2 -6x +3+y^2-2root3 y=12
(x^2+y^2) + 12-6x-2root3 y=12
(x^2+y^2)=6x+2root3 y
put equation (1) value
12=6x+2root3 y----------(2)

step 6:- solve equation (1) and (2)
put equation (1) y=(12-6x)/2root3

x^2+(12-6x)/12=12

=>12 x^2+(12-6x)^2=144

=> 12x^2+36x^2+144-144x=144

=> 48x^2-144x=0

=> x=0,3
so, y=12/2root3=2root3
y=-root3
hence unknown point is (0,2root3) and (3,-root3)