Finding a square in Cartesian coordinate system

[Peter_Hls]

Hi there!

There is a point S(3,2) given in Cartesian coordinate system.
Construct a square ABCD, so that vertex A lies on Y axis and C on X axis where point S is the center of the square.

It is from a high school math book, not providing answer to this Q.
I have tried to solve it in GeoGebra, but I am perplexed and have no idea how to approach this problem.
PS: I am sorry for my English, I don't know how to translate the problem more correctly.

 

[Peter_Hls]

Hello!

I have been solving one problem, but can't find a way to go about it.

There is a point S(3,2) in Cartesian coordinate system.
construct a square ABCD so that vertex A lies on Y-axis and vertex C lies on X-axis and the point S is the center of the square.

Hope you can give me some direction, because I have no idea how to get any more information.

 

[pka]

There is a point S(3,2) given in Cartesian coordinate system.
Construct a square ABCD, so that vertex A lies on Y axis and C on X axis where point S is the center of the square.
It is from a high school math book, not providing answer to this Q.

We use the fact that the diagonals of a square are equal in length, bisect each other at right angles.
So the point $S$ is their intersection. Using the midpoint formula we points $A: (0,4)~\&~B: (6,0)$.
Thus $\overline{AB}$ is one of the diagonals. Can you find the other, the endpoints of which are other vertices.

 

[Dr.Peterson]

Hello!

I have been solving one problem, but can't find a way to go about it.

There is a point S(3,2) in Cartesian coordinate system.
construct a square ABCD so that vertex A lies on Y-axis and vertex C lies on X-axis and the point S is the center of the square.

Hope you can give me some direction, because I have no idea how to get any more information.

Since A and C are opposite points on the square, how are they related to the center, S?

If A is (0, a) and C is (c, 0), write an equation or two expressing that relationship. (There are a couple different ways to do this, but one is very easy.)

If that isn't enough, write back and tell us your answer to my first question, and any work you've done.

(I see you submitted this question twice. pka's answer in the other this [corrected]thread deals mostly with the second step after the part I've suggested.)

 

[pka]

We use the fact that the diagonals of a square are equal in length, bisect each other at right angles.
So the point $S$ is their intersection. Using the midpoint formula we points $A: (0,4)~\&~C: (6,0)$.
Thus $\overline{AC}$ is one of the diagonals. Can you find the other, the endpoints of which are other vertices.

EDIT:It should read:

So the point $S$ is their intersection. Using the midpoint formula we points $A: (0,4)~\&~C: (6,0)$.
Thus $\overline{AC}$ is one of the diagonals. Can you find the other, the endpoints of which are other vertices.

 

[Peter_Hls]

Firstly, I am sorry for submitting this question twice(Later as I opened the forum, the question was not shown in my browser, am really sorry, hope moderator would be willing to move the 2 messages here and delete the redundant one) [done]

So I found out the center is the arithmetic average of the line segment (x1 + x2) / 2 = center's x - axis ...similar to y coordinates.
I found out that a diagonal's length is c**2 = 4**2 + 6**2 ; c = sqrt(52) ....and a side of the square is roughly 5.1 = sqrt(26) (becuase the square comprises of 4 isosceles triangles (45-45-90) .... I just can't don't know how the coordinate's of the vertices of the square relate to each other.

I have tried dozens of sketches, tried to solve it by cos 45° ....but it leads into more problems...
Graphically I found out that B is at (1,-1) and D at (5,5).... But how to get there with math solely?

 

[Dr.Peterson]

Firstly, I am sorry for submitting this question twice(Later as I opened the forum, the question was not shown in my browser, am really sorry, hope moderator would be willing to move the 2 messages here and delete the redundant one) [done]

This often happens when people don't bother to read the post that says "Read Before Posting". Moderation causes your first few posts not to be visible for a while.

So I found out the center is the arithmetic average of the line segment (x1 + x2) / 2 = center's x - axis ...similar to y coordinates.
I found out that a diagonal's length is c**2 = 4**2 + 6**2 ; c = sqrt(52) ....and a side of the square is roughly 5.1 = sqrt(26) (becuase the square comprises of 4 isosceles triangles (45-45-90) .... I just can't don't know how the coordinate's of the vertices of the square relate to each other.

I have tried dozens of sketches, tried to solve it by cos 45° ....but it leads into more problems...
Graphically I found out that B is at (1,-1) and D at (5,5).... But how to get there with math solely?

If (x1 + x2) / 2 = 3 and x1 = 0, what is x2? You didn't write this equation specifically for this problem and solve for x, which is what I suggested. Do the same to find y1.

That will give you points A and C. (I'm assuming you didn't find them, since you never stated what they are, though perhaps your determination of the length of the diagonal implies you did.)

Then you might think about the slope of perpendicular lines to find B and D, since BD is perpendicular to AC. If you were familiar with basic ideas of vectors, this step would be easier. Just start at the center and make the same moves you take to get to A, but at right angles (change the vertical to horizontal and vice versa).

 

[Peter_Hls]

Center point x coordinate = (x1 + x2) / 2 and respectivelly for y
x2 = 2 * S_x - x1 x2 = 2 * 3 - 0 = 6 S_x denotes to x-axis coordinate of the center point
y1 = 2 * S_y - y2 y1 = 2 * 2 - 0 = 4 point A(x1,y1) == A(0,4) point C(x2,y2) == C(6,0)

Thank you all for your help and the direction on what I should learn.
Will post the answer after I will learn about vectors and slopes of lines.

 

[Peter_Hls]

It's messy and mathematically not correct,
but technically vector a refers to the same magnitude everywhere, respectivelly for b
a+b is magnitude of the half of diagonal of the square. -a-b is the same length, but opposite direction.
slopes... well I have some notes, but currently have a hard time from life...
ok so the negative reciprocal of the slope formula gives us the line from B to point D which is 3 point higher and 2 points to the left from the center point of the square...

 

[Dr.Peterson]

I think you've got all the ideas more or less right. Even your crowded picture looks essentially like what I would have shown you, which I call a pinwheel diagram. The perpendicular vector is closely related to the perpendicular slope, as each swaps the role of x and y and changes a sign.