How do I find the coordinates of B and D?

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Kulla_9289

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ABCD is a square. A is the point (-2, 0) and C is the point (6, 4). AC and BD are diagonals of the square, which intersects at M. Find the coordinates at M.

My attempt:
M will be the midpoint of AC because AC and BD are perpendicular to each other. So, M is (2, 2). However, I am no so sure on how to find B and D.
 
Kulla_9289 said:
ABCD is a square. A is the point (-2, 0) and C is the point (6, 4). AC and BD are diagonals of the square, which intersects at M. Find the coordinates at M.

My attempt:
M will be the midpoint of AC because AC and BD are perpendicular to each other. So, M is (2, 2). However, I am no so sure on how to find B and D.
Click to expand...
This is easier to describe if you know anything about vectors; but if you imagine (or draw) a right triangle with MC as its hypotenuse and with legs parallel to the axis, you can see what will happen if you rotate that triangle 90 degrees, keeping the vertex at M fixed. That will give you B or D.
 
What formula is this [imath](\frac{x_1+x_3+y_3-y_1}{2}, \frac{x_1-x_3+y_1+y_3}{2})[/imath]?
 
Draw a diagram with AC drawn in. Then draw in the other diagonal BD.

Imagine walking from Point M to Point C. How may steps east and then how many steps north? Pencil that in.

Can you then see how to get from M to B - ie how many steps north and then how many steps west?
 
Kulla_9289 said:
What formula is this [imath](\frac{x_1+x_3+y_3-y_1}{2}, \frac{x_1-x_3+y_1+y_3}{2})[/imath]?
Click to expand...
I have no idea what that "formula" is; I have never come across it before. Where did you get it?
Kulla_9289 said:
ABCD is a square. A is the point (-2, 0) and C is the point (6, 4). AC and BD are diagonals of the square, which intersects at M. Find the coordinates at M.

My attempt:
M will be the midpoint of AC because AC and BD are perpendicular to each other. So, M is (2, 2). However, I am no so sure on how to find B and D.
Click to expand...
A geometrical solution to your problem should be fairly straightforward.
Start by plotting your two given points (A & C) then draw the diagonal of the square between them.
Can you "see" (or calculate) the gradient of AC?
You know BD is perpendicular to AC so do you know what the gradient of BD will be (given the gradient of AC)?
(mAC × mBD = -1)
Since BD passes through AC at M that should allow you to draw in the (dotted?) line that B & D lie on.
Consider the point P (for Pythagoras?). See pic below.
You could calculate the length of MC but it may be easier just to look at its 'displacement' in the x- & y- directions.
Since AC bisects BD, then AM, BM, CM & DM are all the same length.
That should now allow you to figure out where B & D lie.

Square.png​

Hope that helps. ?

NB: Please come back and show us your answers.
 
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B(4, -2). By the way, why does rotating 90 degrees work? What's the concept behind this?
 
Kulla_9289 said:
B(4, -2). By the way, why does rotating 90 degrees work? What's the concept behind this?
Click to expand...
No! B isn't (4, -2) though that is one of the vertices (corners).
Look at the triangle MCP (in my diagram).
If you keep the point M where it is and rotate that triangle through 90° (anti-clockwise) then the point C will move to B, then A, then D and then back to its original position (C).
Can you not visualize that?

The hypotenuse of the triangle, MC, becomes MB, then MA, then MD then, finally, rests back in its original position, MC, again.
(Which are all the same length, as I pointed out to you earlier.)
 
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For the avoidance of any possible confusion, the coordinates are correct but you have assigned (given) them to the wrong points; swap 'em round. ?
 
I would solve this by first finding M:

\(\displaystyle M=\left(\frac{-2+6}{2},\frac{0+4}{2}\right)=(2,2)\)

Now, we can find the slope of AC:

\(\displaystyle m=\frac{4-0}{6+2}=\frac{1}{2}\)

And the square of the length of AM:

\(\displaystyle r^2=(2+2)^2+(2-0)^2=20\)

And so B and D will be the solutions to the system:

\(\displaystyle y-2=-2(x-2)\)

\(\displaystyle (x-2)^2+(y-2)^2=20\)

From this we find:

\(\displaystyle x-2=\pm2\)

\(\displaystyle y-2=\mp4\)

It is easy then to determine:

\(\displaystyle B=(2-2,2+4)=(0,6)\)

\(\displaystyle D=(2+2,2-4)=(4,-2)\)
 
The Highlander said:
No! They are the wrong way round!!
Your square would then be ADCB when it should (by convention) be ABCD.
Nearly there.....
Have one last go, eh? ?
Click to expand...
Are you assuming the square has to be named in clockwise order? Nothing in the problem requires that.
Kulla_9289 said:
What formula is this [imath](\frac{x_1+x_3+y_3-y_1}{2}, \frac{x_1-x_3+y_1+y_3}{2})[/imath]?
Click to expand...
That would appear to be a formula for one of the other vertices, which can be derived from the methods you have been told. Where did you get it?
Kulla_9289 said:
B(4,-2) and D(0,6). Are these not correct?
Click to expand...
Yes, they are, if you either prefer, or accept, the counterclockwise naming (as I do). Your formula gives B, and swapping the input points gives D.
Kulla_9289 said:
@Harry_the_cat in vector form (+4,+2) displacement from m to c
Click to expand...
Since you do know about vectors, have you learned anything about rotating vectors, or about perpendicular vectors?
 
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