In the following examples , can the segment joining the given points from a triangle ? If triangle is formed , state the type of the triangle considering sides of the triangle.
(iii) A `(sqrt(2) ,sqrt(2)) , B (-sqrt(2) ,-sqrt(2) ), C (-sqrt(6) ,sqrt(6) )`

Explanation

`A(sqrt(2),sqrt(2)),B(-sqrt(2),-sqrt(2))` and `C(-sqrt(6),sqrt(6))`
By distance formula,
`AB=sqrt((-sqrt(2)-sqrt(2))^(2)+(-sqrt(2)-sqrt(2))^(2))`
`=sqrt((-2sqrt(2))^(2)+(-2sqrt(2))^(2))`
`=sqrt(8+8)`
`=sqrt(16)`
`=4`
`BC=sqrt([-sqrt(6)-(-sqrt(2))]^(2)+[sqrt(6)-(-sqrt(2))]^(2))`
`=sqrt((-sqrt(6)+sqrt(2))^(2)+(sqrt(6)+sqrt(2))^(2))`
`=sqrt((-sqrt(6)^(2)+2(-sqrt(6))(sqrt(2))+(sqrt(2))^(2)+(sqrt(6))^(2)+2(sqrt(6))(sqrt(2))+(sqrt(2))^(2))`
`=sqrt(6-2sqrt(12)+2+6+2sqrt(12)+2)`
`=sqrt(16)`
`=4`
`AC=sqrt((-sqrt(6)-sqrt(2))^(2)+(sqrt(6)-sqrt(2))^(2))`
`:.AC=sqrt((-sqrt(6))^(2)-2(-sqrt(6))(sqrt(2))(sqrt(2))^(2)+(sqrt(6))^(2)-2(sqrt(6))(sqrt(2))+(sqrt(2))^(2))`
`AC=sqrt(6+2sqrt(12)+2+6-2sqrt(12)+2)`
`AC=sqrt(16)`
`:.AC=4`
`:.AB=BC=AC=4`
Here `AB+BCgtAC,BC+ACgtAB` and `AB+ACgtBC`
`:.DeltaABC` is formed and since its all sides are equal it is an equilateral triangle.
The segment joining `A(sqrt(2),sqrt(2)), B(-sqrt(2),-sqrt(2))` and `C(-sqrt(6),sqrt(6))` form an equilateral triangle.