Question

The endpoints of a diagonal of a square are (5,-1) and (-7,3). Find the coordinates of the two other vertices.

A little help please? :)

Answer 1

np. First get the distance between the 2 points. That distance will be the same for the other two points. For the original 2 points, get slope, then get negative reciprocal for perpendicularity. Get midpoint of original points. For last part, get two points for which this midpoint is also midpoint of the new perpendicular line. This is a highly condensed explanation and we now need to expand this and take it step-by-step.

Answer 2

Do you know how to get the distance between the 2 original points?

Answer 3

using the distance formula ^_^

Answer 4

Good! Can you get the slope?

Answer 5

Yeah. m = y-y1/x-x1 :)

Answer 6

If you forgot, it's \[\frac{ y _{1}-y _{2} }{ x _{1}-x _{2} } = m\] Okay, looks like you knows that, Great!

Answer 7

The diagonal of the other points will be perpendicular, so get a new slope for the upcoming next 2 points for that line.

Answer 8

Negative reciprocal, but I would guess that you know that, too.

Answer 9

ahh, negative reciprocal because its perpendicular :) yes yes ^^

Answer 10

With the midpoint of the given 2 points (which will also be the midpoint of the new 2 points) and with the negative reciprocal slope, the new points will be on that line, so you will use the slope formula and midpoint distance twice, once for each new point.

Answer 11

You might have done so already, but making a drawing will help a bit, if you need to see it and if you have any trouble just visualizing.

Answer 12

woo. *still trying* ^^

Answer 13

np. At this point, it's just algebra. I think you have the conceptual underpinings down now.

Answer 14

haha :) even my classmates are confused. XD

Answer 15

Another way to have done this problem is to use distance formula, get half of the distance for the radius of a circle with the midpoint of the two points as the center of that circle. Still uses negative reciprocal. It's basically the same approach with a slightly different conceptualization.

Answer 16

do i have to equate the two equations formed?

Answer 17

my head my head >< worrrrkkk

Answer 18

\[y _{1} - y _{midpoint} = (-1/m)(x _{1} - x _{midpoint})\]Yes, equate values, using the above formula rearranged and "half the distance" formula using midpoint and one of the new points.

Answer 19

Yes, it's a lot of algebra, but it's straight-forward now. I have to believe you can easily follow paper-and-pencil constructions steps to guide your equations. It's a brute-force problem!

Answer 20

I forgot to ask, you didn't have any trouble getting that midpoint, did you? Do you need help with that?

Answer 21

(-1,1)?

Answer 22

Yes, and I can tell that you are fairly advanced and a good student. You're probably taking Analytic Geometry right now?

Answer 23

Ahm. i think so. haha. i'm only 14

Answer 24

honestly, it's already 2 am here and i have to make my head work ^^ahahaa

Answer 25

im sorry if i couldn't get it that fast. :/

Answer 26

This is an advanced problem for Algebra I. I'm a little surprised that a problem of this complexity was given to you for age 14.

Anyway, once you equate the slope formula I wrote out for you as my latest formula and the distance, you basically substitute the right side into the distance formula, so you are getting the value of the x-coordinates of the new points (in the adjusted distance formula). Once you get the x-coordinates, you can go back to the slope formula to get the y's.

And no problem on doing this fast or not. It's not really an easy problem. Several steps and a lot of conceptualizing.

Answer 27

i think i got the one missing vertex. haha!
(-3, -5)?

Answer 28

i also got (1,7) -_-

Answer 29

@tcarroll010 ? :)

Answer 30

Still here. As I was posting, I was thinking conceptually and had not done the computations. I would have to get paper and pencil and do them. You want me to check your answers?

Answer 31

yes yes :)

Answer 32

For the original 2 points, D = sqrt(160), so d is half which is distance from midpoint to any of the 4 points so d= [sqrt(160)]/2. Midpoint is (-1, 1) which already now by now. The original slope is 4/(-12) so the slope of the new points is m=3. Still typing and figuring.

Answer 33

I'll be waiting :) haha!

Answer 34

m = 3 = (y-1)/(x+1), so y-1 = 3/(x+1). d = [(x+1)^2 + (y-1)^2]^(1/2) = [sqrt(160)]/2. Check my algebra for accuracy. Keep me on track.

Answer 35

Substituting, [sqrt(160)]/2 = [(x+1)^2 + (3/(x+1))^2]^(1/2) Now, checkpoint: did you do this, or at least, does this appear sensible to you at this point?

Answer 36

sensible :)
haha. you gave me ideas and tried and tried

Answer 37

and I*

Answer 38

in order to check answers, all sides must have the same distance right? :) :)

Answer 39

Continuing, squaring both sides: 40 = (x+1)^2 + [3/(x+1)]^2 40(x+1)^2 = (x+1)^4 + 9. Yes, all sides must be same distance, but also, the diagonals distance have to be equal to each other, and that's why we have to go this "long way".

Answer 40

ohh, i got \[4\sqrt{10}\] as distance of the diagonals ^_^

Answer 41

Yes, that's what I call D. And d = D/2 = distance from midpoint to any of the four points.

Also, I'm pretty confident in the steps I am taking and that last equation I derived which is a quadratic in (x+1)^2. You can substitute n = (x+1)^2 and solve the quadratic in "n", and then once you get "n", you can substitute back for "x". I have to go pick up someone at a hospital and will be gone for a bit. I know it's late where you are, but you have the whole methodology now and most of the equations. You can either wait for me to come back, or you can probably finish this out on your own now. Sorry I have to go.

Answer 42

Ahm yeah it's okay :)
Btw, thank you for your help! :D

Answer 43

I just got back from the hospital, and [sqrt(160)]/2 = [(x+1)^2 + (3/(x+1))^2]^(1/2) gives x = {1, -3}. Using these in the slope formula will give you your 2 new points (-3, -5) and (1, 7), so you had the right new points. Good job! Sorry for the interruption.

Answer 44

Ahm again, it's okay :)
Thanks again for your help!! :D