Question

Prove that the points 2, -1+i√3, -1-i√3 for a equilateral triangle on the argand plane.

Find the length of a side of this trangle?

Answer 1

this is just a quick addition to the superb posting by "hamza0100" above

well, indeed, in the argand or imaginary plane, for those values above we have the coordinates of A(2 , 0) , B(-1 √3) and C(-1 , -√3), let' use the distance formula for those fellows

`~\hfill \stackrel{\textit{\large distance between 2 points}}{d = √(( x_2- x_1)^2 + ( y_2- y_1)^2)}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{2}~,~\stackrel{y_1}{0})\qquad B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{√(3)}) ~\hfill AB=\sqrt{(~~ -1- 2~~)^2 + (~~ √(3)- 0~~)^2} \\\\\\ ~\hfill AB=\sqrt{( -3)^2 + ( √(3))^2} \implies \boxed{AB=√( 12 )}`

`B(\stackrel{x_1}{-1}~,~\stackrel{y_1}{√(3)})\qquad C(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-√(3)}) \\\\\\ BC=\sqrt{(~~ -1- (-1)~~)^2 + (~~ -√(3)- √(3)~~)^2} \\\\\\ ~\hfill BC=\sqrt{( 0)^2 + ( -2√(3))^2} \implies \boxed{BC=√( 12 )}`

`C(\stackrel{x_1}{-1}~,~\stackrel{y_1}{-√(3)})\qquad A(\stackrel{x_2}{2}~,~\stackrel{y_2}{0}) ~\hfill CA=\sqrt{(~~ 2- (-1)~~)^2 + (~~ 0- (-√(3))~~)^2} \\\\\\ ~\hfill CA=\sqrt{( 3)^2 + (-√(3))^2} \implies \boxed{CA=√( 12 )} \\\\[-0.35em] ~\dotfill\\\\ AB=BC=CA=√(12)\implies 2√(3)\hspace{5em}\qquad equilateral\textit{\LARGE \checkmark}`

Answer 2

The lengths are equal so the triangle is equilateral

Explanation:

We can write the points as follows,

(2,0), (-1,
`√(3)`) (-1,-
`√(3)`)

now if it is an equilateral triangle, all side lengths must be equal

first we compute the sides(vectors)

(2-(-1),-
`√(3)`) = (3,-
`√(3)`) = side 1

(2-(-1),
`√(3)`) = (3,
`√(3)`) = side 2

(-1+1,
`√(3)`+
`√(3)`) = (0,2
`√(3)`) = side 3

now we compute the lengths of the sides using pythagoras theorem

(3)^2 + (-
`√(3)`)^2 = (length of side 1)^2 = 9 + 3 = 12

similarly, (3)^2 + (
`√(3)`)^2 = 12 = Length of side 2 squared

and,( 2
`√(3)`)^2 = length of side 3 squared = 12

since the squares are equal, so the lengths must also be equal

so the triangle is equilateral