vertices of a square, coordinate geometry and differentiation

The origin O and a point B(p, q) are opposite vertices of the square OABC. Find the coordinates of the point A and C

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

So I’m stuck on the first part of this problem and haven’t tried to tackle the second part yet, but I posted it anyway in case I have difficulties with it once I have solved the first part.

For the first question I observed that I have 4 unknowns, the two coordinates of each point. So I figured I should determine 4 equations that would help me find these; I used the following remarks to define these equations:

The gradient of AC will be perpendicular to the gradient of OB,

The gradient of OA will be perpendicular to the gradient of AB

The gradient of OC will be perpendicular to the gradient of CB,

The distance between A and C will be equal to the distance between O and B

However I’m not sure that this is the right (or at least, most effective) way to proceed as I’ve ended up with a mess of equations, often ending up quadratic, that appear to be leading me nowhere. Am I on the right track, have I used the wrong equations or am I just using a totally incorrect method?

Re: vertices of a square, coordinate geometry and differentiation

https://lh4.googleusercontent.com/-o...800/square.png

Note that the result of counterclockwise rotation by 90 degrees of a vector (p, q) is (-q, p), and the result of clockwise rotation by 90 degrees of (p, q) is (q, -p).

Re: vertices of a square, coordinate geometry and differentiation

Quote:

Originally Posted by

**furor celtica** The origin O and a point B(p, q) are opposite vertices of the square OABC. Find the coordinates of the point A and C

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

So I’m stuck on the first part of this problem and haven’t tried to tackle the second part yet, but I posted it anyway in case I have difficulties with it once I have solved the first part.

For the first question I observed that I have 4 unknowns, the two coordinates of each point. So I figured I should determine 4 equations that would help me find these; I used the following remarks to define these equations:

The gradient of AC will be perpendicular to the gradient of OB,

The gradient of OA will be perpendicular to the gradient of AB

The gradient of OC will be perpendicular to the gradient of CB,

The distance between A and C will be equal to the distance between O and B

However I’m not sure that this is the right (or at least, most effective) way to proceed as I’ve ended up with a mess of equations, often ending up quadratic, that appear to be leading me nowhere. Am I on the right track, have I used the wrong equations or am I just using a totally incorrect method?

Here are my thoughts to this very ambitious question:

1. B is located on a circle line. That means you'll get infinitely many squares which all satisfy the given conditions.

2. The midpoint of the square is $\displaystyle M\left(\frac p2 \ , \ \frac q2 \right)$

The line AC passes through M and has the slope (=gradient) $\displaystyle m = -\frac pq$

It therefore has the equation:

$\displaystyle y-\frac q2 = -\frac pq \left( x - \frac p2 \right)$

3. The vertices of the square are placed on a circle around M with $\displaystyle r = \frac12 \cdot \sqrt{p^2+q^2}$

4. Calculate the coordinates of the points of intersection of the circle and the line AC. The calculations will be very difficult! Good luck!

Re: vertices of a square, coordinate geometry and differentiation

Quote:

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

For this you need $\displaystyle tan(\theta+ \phi)=\frac{tan(\theta)+ tan(\phi)}{1+ tan(\theta)tan(\phi)}$.

Re: vertices of a square, coordinate geometry and differentiation

hi HallsofIvy

shouldn't it be 1-tan(theta)tan(fi) in the denominator?

Re: vertices of a square, coordinate geometry and differentiation

the problem is, i don't know how to formualte the euqation of a circle (if that even exists) and i'm learning about vectors in the chapter immediately after this one, so niether of the methods proposed above apply to my level. any other suggestions? this can't be too hard.

Re: vertices of a square, coordinate geometry and differentiation

Quote:

Originally Posted by

**furor celtica** the problem is, i don't know how to formualte the euqation of a circle (if that even exists) and i'm learning about vectors in the chapter immediately after this one

So, we have the following claim.

The result of turning a point (p, q) counterclockwise by 90 degrees around the origin is (-q, p).

You can prove it by dropping perpendiculars from the two points to the horizontal axis and proving that the resulting triangles are equal.

You can also prove it using trigonometry. Let $\displaystyle \alpha$ be the angle between the horizontal axis and the line from the origin to (p, q). Let also d be the distance from the origin to (p, q), i.e., $\displaystyle d=\sqrt{p^2+q^2}$. Then $\displaystyle p = d\cos\alpha$ and $\displaystyle q=d\sin\alpha$. The coordinates of the rotated point are $\displaystyle (d\cos(\alpha+\pi/2),d\sin(\alpha+\pi/2))=(-d\sin\alpha,d\cos\alpha)=(-q,p)$.