If ` A(2,2), B(-2, -2) , C(-2sqrt(3), 2sqrt(3)) and D(-4-2sqrt(3), 4+2sqrt(3))` are the co-ordinates of 4 points. What can be said about these four points ?

If ` A(2,2), B(-2, -2) , C(-2sqrt(3), 2sqrt(3)) and D(-4-2sqrt(3), 4+2

1.	If ` A(2,2), B(-2, -2) , C(-2sqrt(3), 2sqrt(3)) and D(-4-2sqrt(3), 4+2sqrt(3))` are the co-ordinates of 4 points. What can be said about these four points ?
Answer» `" "AB=sqrt((-2-2)^(2)+(-2-2)^(2))=4sqrt(2)` units `" "BC=sqrt((-2+2sqrt(3))^(2)+(-2-2sqrt(3))^(2))` `" "=sqrt (4+12-8sqrt(3)+4+12+8sqrt(3))=4sqrt(2) ` units `" "CD=sqrt((-2sqrt3+4+2sqrt(3))^(2)+(2sqrt(3)-4-2sqrt(3))^(2))` ` " " =sqrt(16+16)=4sqrt(2)` units `" "AC=sqrt((2+2sqrt(3))^(2)+(2- 2sqrt(3))^(2))` `" "=sqrt(4+12+8sqrt(3)+4+12-8sqrt(3))=4sqrt(2)` units ` " "AD=sqrt((2+4+2sqrt(3))^(2)+(2-4-2sqrt(3))^(2))` `" "=sqrt(36+12+24sqrt(3)+12+4+8sqrt(3))=sqrt(64+32sqrt(3))` units `" "BD=sqrt((-2+4+2sqrt(3))^(2)+(-2-4-2sqrt3)^(2))` `" "=sqrt(4+12+8sqrt(3)+36+12 +24sqrt(3))=sqrt(64+32sqrt( 3))` units Here, `AB =BC=CD=AC` and also, `AD=BD` So, in first view it seems to be the vertices of a square. `" "BUT" "` Here, ` AB, BC, CD and DA` are not equal. (order of `A, B, C and D` must be cyclic in case of square). Also AD and BD are equal but they cannot be the diagonals. So, they do not form a square. Actually, A, B and D lie on a circle with C as the centre (as CA=CB=CD i.e., C is equidistant from A, B and D).