Two vertices of an equilateral triangle are (a,-a) and (-a,a). Find the third.

Hello!

An equilateral triangle, by definition, has all three sides of equal lengths.

For two points with the coordinates `(x_1,y_1)` and `(x_2,y_2)` the distance between them is

`d=sqrt((x_1-x_2)^2+(y_1-y_2)^2).`

The length of the given side is `sqrt((2a)^2+(-2a)^2)=2|a|sqrt(2).`

Denote the third point coordinates as `x` and `y,` then the lengths of the other side are

`d_1^2=sqrt((x-a)^2+(y+a)^2)` and `d_2^2=sqrt((x+a)^2+(y-a)^2).`

And `d_1=d_2=2|a|sqrt(2).` This gives us the system of two equations with two variables.

Both sides of both equations are nonnegative, so we can square both equations and will obtain

`8a^2=x^2-2ax+a^2+y^2+2ay+a^2,`
`8a^2=x^2+2ax+a^2+y^2-2ay+a^2.`

Subtract the first from the second and obtain

`0=4ax-4ay,` or `x=y` (if `a!=0`).

Now substitute this into the first equation and obtain

`8a^2=2x^2+2a^2,` or `x^2=3a^2,` or `x=+-|a|sqrt(3).`

So for `a!=0` there are two solutions, `(|a|sqrt(3),|a|sqrt(3))` and `(-|a|sqrt(3),-|a|sqrt(3)).`

For `a=0` the only solution is `(0,0).`