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 formula for the line on which A and C lie...

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 formula for the line on which A and C lie (y=-px/q+p²/2q+q/2), and the distance that they are from the midpoint which i have dubbed "M" for reference sakes (M: (p/2,q/2) distance: (sqrt(p²+q²))/2

I have no idea how to combine this data to find A and C, but i know it is all relevant information
thanks in advance

Asked on by dark0flame

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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).

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