Finding vertices given opposite vertices

[eddybob123]

Hi everyone, is there a way of finding the other two vertices of a square on the coordinate system given that two opposite vertices are already there? I already attempted this problem with two methods: with Pythagoras and with circles. So, is there a general formula or not?

 

[tiny-tim]

hi eddybob123!

hint: first find the centre

then … ​

 

[eddybob123]

I tried that too. I ended up with two equations with two variables including their squares, and I didn't know how to solve for them, even with the quadratic formula.

 

[tiny-tim]

but once you've found the centre, all you need do is to find the line through it at right-angles to the given diagonal

 

[eddybob123]

Wait, the center of what? And can you define what you mean more precisely, maybe provide a diagram? I think I have also tried that, but I might have overlooked it because it seemed so easy.

 

[tiny-tim]

the center of the square (and of the given diagonal)!

And can you define what you mean more precisely, maybe provide a diagram?

erm, I've a better idea …

you define what you mean more precisely, and maybe provide a diagram ​

 

[eddybob123]

The two opposite vertices are labeled (a,b) and (c,d). The points labeled (e,f) and (g,h) are the points I wish to find (in terms of a, b, c, and d). Can you just give me your approach to this problem?

 

[tiny-tim]

The two opposite vertices are labeled (a,b) and (c,d). The points labeled (e,f) and (g,h) are the points I wish to find (in terms of a, b, c, and d).

ok, then the centre dot is at ( , ) ?

and the slope of the second diagonal is ?

 

[eddybob123]

The center is at the midpoint of the line connecting (a,b) and (c,d), that is, by the distance formula, (sqrt((c-a)²+(d-b)²))/2. The slope of the perpendicular line is (a-c)/(d-b). As I told you, I have attempted this method before, but the equations I got were terribly complicated to solve for.

 

[tiny-tim]

The center is at the midpoint of the line connecting (a,b) and (c,d)

so what are its coordinates?

 

[eddybob123]

I am pretty sure it is ((c+a)/2,(d+b)/2), but how does that concern finding the other vertices?

 

[Curious3141]

Hi everyone, is there a way of finding the other two vertices of a square on the coordinate system given that two opposite vertices are already there? I already attempted this problem with two methods: with Pythagoras and with circles. So, is there a general formula or not?

Have you learned complex numbers? I find that the problem becomes almost trivially easy when you use these.

You can also use vectors, but I just find the problem easier to do with complex numbers. They're basically equivalent anyway.

Using coordinate geometry is quite the slog, though.

If you're familiar with complex numbers, I can take you through the steps. So let us know.

 

[tiny-tim]

(just got up :zzz:)

I am pretty sure it is ((c+a)/2,(d+b)/2)

that's right!

isn't it obvious? the centre is the average of the two endpoints, so it must be half of one and half of the other

here's the start of the proof for the position of the centre C of AB, for you to finish …

vector AC = vector CB, so C - A = B - C, so C = … ? ​

ok, now if the vector OP has coordinates (p,q), what are the coordinates of the perpendicular vector of the same length?

 

[HallsofIvy]

Here's another way: the line through (a, b) and (c, d) has slope (d- b)/(c- a) so the line perpendicular to that has slope (a- c)/(d- b) (-1 times the reciprocal). The line with that slope through the center point, ((a+c)/2, (b+ d)/2), is given by y= (a-c)/(d-b)(x- (a+c)/2)+ (b+d)/2. The circle with center at ((a+c)/2, (b+d)/2) and radius \sqrt{(a-c)^2+(d- b)^2}/2, and so passing through all four vertices of the square, is given by (x- (a+c)/2)^2+ (y- (b+ d)/2)^2= ((a- c)^2+ (d- b)^2)/4

If you replace y in that last equation with (a-c)/(b-d)(x- (a+c)/2)+ (b+d)/2, you get a quadratic equation in x to solve for the x coordinates of the other two vertices.