Assuming A,B,C,D taken in the clockwise direction. A = (12,2) in first quadrant => D in first quadrant. Let C = (0,c) and B = (b,0).
Now, drop a perpendicular from A to x-axis, call this point E. Then, triangles COB and BEA are congurent. O is the origin. Thus, OC = BE and AE = OB. => A = (b+c,b).
(b+c,b) = (12,2) => b = 2, and c = 10.
D can be found using any of the many properties of a square, e.g. diagonals of a square bisect. if D = (x,y) => (x+b,y) = (b+c,b+c). => (x,y) = (c,b+c) = (10,12).
Curiosity: the center of the square always stays on the line y = x! Prashanth. PS: Kudos Vamsee!
A(12,2)
ReplyDeleteB(2,0)
C(0,10)
D(10,12)
fantastic !! if u could eloborate the solution might it be useful to many.. Vamsee
ReplyDeleteGeometric proof:
ReplyDeleteAssuming A,B,C,D taken in the clockwise direction.
A = (12,2) in first quadrant => D in first quadrant.
Let C = (0,c) and B = (b,0).
Now, drop a perpendicular from A to x-axis, call this point E.
Then, triangles COB and BEA are congurent. O is the origin.
Thus, OC = BE and AE = OB. => A = (b+c,b).
(b+c,b) = (12,2) => b = 2, and c = 10.
D can be found using any of the many properties of a square, e.g. diagonals of a square bisect.
if D = (x,y) => (x+b,y) = (b+c,b+c). => (x,y) = (c,b+c) = (10,12).
Curiosity: the center of the square always stays on the line y = x!
Prashanth.
PS: Kudos Vamsee!
YOU DID IT !!
ReplyDelete