the origin O and a point B(p,q) are opposite vertices of the square OABC. Find the coordinates of the points A and C.got as far as finding the...

If we'll join the opposite vertices of the square we'll
get the diagonal of the square, which represents also the hypotenuse of the right angle
triangle OAB, whose right angle is A.

Since the lengths of
the sides of the square are equal, let the sides OA and AB be
x.

We'll apply Pythagorean theorem, in right triangle OAB, 
to get OB:

`x^(2)` + `x^(2)` =
`OB^(2)`

2`x^(2)` =
`OB^(2)`

OB =
x`sqrt(2)`(1)

But, we can calculate the length of the
hypotenuse OB, using the distance formula:

OB =
`sqrt((xB-xO)^(2) +(yB-yO)^(2))`

OB = `sqrt(p^(2)+q^(2))`
(2)

Now, we can equate (1) and
(2)

x`sqrt(2)` =
`sqrt(p^(2)+q^(2))`

x = `sqrt(2(p^(2) + q^(2)))`
/2

Since we know the length of each side of the square, now
we can find the coordinates of the vertices A and C, using the distance
formula:

OA = `sqrt((xA-xO)^(2) + (yA -
yO)^(2))`

OA = `sqrt(xA^(2) +
yA^(2))`(3)

But OA = `sqrt(2(p^(2) + q^(2)))` /2
(4)

We'll equate (3) and
(4):

`xA^(2)` + `yA^(2)` = (`p^(2)` + `q^(2)`
)/2

We also can calculate the length of the side
BA:

BA = `sqrt((p-xA)^(2) +
(q-yA)^(2))`(5)

We'll equate (5) and
(4):

`(p-xA)^(2)` + `(q-yA)^(2)` = (`p^(2)` + `q^(2)`
)/2

We'll expand the binomials from the left
side:

`p^(2)` - 2pxA + `xA^(2)` + `q^(2)` - 2qyA + `yA^(2)`
= (`p^(2)` + `q^(2)` )/2

But `xA^(2)` + `yA^(2)` = (`p^(2)`
+ `q^(2)` )/2

2(`xA^(2)` + `yA^(2)` ) - 2(pxA  + qyA) =
0

xA(xA - p) = yA(yA - q)

The
length of diagonal OB being equal with the length of diagonal AC, therefore, we cand
determine the coordinates of the vertex
C.

Therefore, the relation between
coordinates of vertices A and B is: xA(xA - p) = yA(yA -
q).