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coordinate geometry

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.
A line l has gradient q/p. Find possible values for the gradient of a line at 45° to l.

I realise OB gradient is q/p and CA is the normal to it (or perpendicular) so AB's gradient must be -p/q. From here i guess i could use the lines formula but it doent work out to nearly the answer... please help
Reply 1
Original post by Jmun
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.
A line l has gradient q/p. Find possible values for the gradient of a line at 45° to l.

I realise OB gradient is q/p and CA is the normal to it (or perpendicular) so AB's gradient must be -p/q. From here i guess i could use the lines formula but it doent work out to nearly the answer... please help


Which bit are you trying to solve here - the first or second question?

Unless I've misunderstood the question, you should be able to write down the answers for A and C for the first part.

The second part is a bit more complicated - what have you tried so far?
Reply 2
Original post by ghostwalker
Unless I've misunderstood, the first part is a bit of a pig.


You could well be right :biggrin:

(I only glanced at the question and didn't take in exactly what was required - my brain is winding down for the evening now!)
Original post by davros
You could well be right :biggrin:

(I only glanced at the question and didn't take in exactly what was required - my brain is winding down for the evening now!)


Don't worry - it's my brain that wasn't functioning. Just ignore me.
Reply 4
I'm trying to solve the first part; the answer is (1/2(p-q),1/2(p+q)), (1/2(p+q),1/2(q-p)) but i have no idea how they get this!!
Original post by Jmun
I'm trying to solve the first part; the answer is (1/2(p-q),1/2(p+q)), (1/2(p+q),1/2(q-p)) but i have no idea how they get this!!


Do you know how to do a rotation by 90 degrees about the origin, in terms of coordinates?
(edited 8 years ago)
Reply 6
Would it work to find the length of MB and point M coordinates and use the inverse of the gradient of OB and use that to find A and C? I don't think im on the right track though
Original post by Jmun
Would it work to find the length of MB and point M coordinates and use the inverse of the gradient of OB and use that to find A and C? I don't think im on the right track though


I presume that M is the centre of the square - midpoint of OB.

Yes, you could find the eqn of the other diagonal through M, and then use distance to O or B, having worked out the length of a side.

However, if the answer to my previous post was yes, then there is a lot simpler method.
Reply 8
Okay then could you perhaps show me the simpler method please? I'm unsure of how to do a rotation by 90 degrees about the origin.
Original post by Jmun
Okay then could you perhaps show me the simpler method please? I'm unsure of how to do a rotation by 90 degrees about the origin.


This is only really useful if you've covered some of this material already - otherwise ignore it.

A rotation of 90 degrees counterclockwise maps the point (x,y) to the point (-y,x)

Given the two points O,B, if we could rotate them 90degrees about the centre of the square we would get the points A,C.

Since we can easily rotate about the origin, we shift the origin to the centre of the square. perform the rotation, and then shift the origin back. The net effect is to rotate the square about its centre.

I have deliberately not put all the details in.
Reply 10
great thanks!

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