# Thread: vertices of a square, coordinate geometry and differentiation

1. ## 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?

2. Originally Posted by red and white kop!
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.
Assuming that the square has a side parallel to the two x-axis and $p>0~\&~q>0$ then
$A0,q)~\&~Cp,0).$

Is the second question part of the first?

3. The square doesn't necessarily have a side parallel to the axes. All information is given in the question. This could be any square.

4. Originally Posted by red and white kop!
The square doesn't necessarily have a side parallel to the axes. All information is given in the question. This could be any square.
In that case, find the points of intersection of the line $y-\frac{q}{2}=\frac{-p}{q}\left(x-\frac{p}{2}\right)$ and the circle $\left(x-\frac{p}{2}\right)^2+\left(y-\frac{q}{2}\right)^2=\frac{p^2}{4}+\frac{q^2}{4}$.

Those two points are $A~\&~C$.

5. Hello, red and white kop!

Did you make a sketch?

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

(a) Find the coordinates of the point A and C.

(b) A line L has gradient q/p. Find possible values for the gradient of a line at 45° to L.
Code:

|
|       B
|       o(p,q)
|      *
|     *
|  M *
|   o(p/2,q/2)
|  *
| *
|*
- - o - - - - - -
O|

We are given points $O(0,0)\text{ and }B(p,q)$
. . and their midpoint $M\left(\frac{p}{2},\,\frac{q}{2}\right)$

Code:
 |
|       B
|       o(p,q)
|      *
|     * ↑
|  M *
|   o → +
|  *
| *
|*
- - o - - - - - -
O|
To move from $M$ to $B$, we move $\frac{p}{2}$ right and $\frac{q}{2}$ up.

Code:
  |
|       B
|       o(p,q)
|      *
|     *
|  M *
|   o → +
|  *    ↓
| *     ↓
|*      o A
- - o - - - - -
O|
To move from $M$ to $A$, we must move $\frac{q}{2}$ right and $\frac{p}{2}$ down.
Do you see why?
. . Hence, $A$ is at $\left(\frac{p+q}{2},\,\frac{p-q}{2}\right)$

Similarly, we find that $C$ is at $\left(\frac{p-q}{2},\,\frac{p+2}{2}\right)$

6. Originally Posted by soroban
Hello, red and white kop!

Did you make a sketch?

Code:

|
|       B
|       o(p,q)
|      *
|     *
|  M *
|   o(p/2,q/2)
|  *
| *
|*
- - o - - - - - -
O|

We are given points $O(0,0)\text{ and }B(p,q)$
. . and their midpoint $M\left(\frac{p}{2},\,\frac{q}{2}\right)$

Code:
 |
|       B
|       o(p,q)
|      *
|     * ↑
|  M *
|   o → +
|  *
| *
|*
- - o - - - - - -
O|
To move from $M$ to $B$, we move $\frac{p}{2}$ right and $\frac{q}{2}$ up.

Code:
  |
|       B
|       o(p,q)
|      *
|     *
|  M *
|   o → +
|  *    ↓
| *     ↓
|*      o A
- - o - - - - -
O|
To move from $M$ to $A$, we must move $\frac{q}{2}$ right and $\frac{p}{2}$ down.
Do you see why?
. . Hence, $A$ is at $\left(\frac{p+q}{2},\,\frac{p-q}{2}\right)$

Similarly, we find that $C$ is at $\left(\frac{p-q}{2},\,\frac{p+2}{2}\right)$
Soroban, thanks for your time, your method seems to be the most applicable my level of comprehension and the section I’m studying.
However, when you say ‘do you see why’, I do not: it would help if you formulated your answer more algebraically and less with phrases and diagrams. That would help me understand your logic a lot better. Also your answer seems to be incorrect, as the answers given in the textbook are (0.5 (p-q), 0.5(p+q)) and (0.5(p+q), 0.5(q-p)); but I’ve discovered several errors in this textbook’s answer key recently so I’m not sure if you are wrong or the textbook is. I’d be thankful if you could keep helping me out until I’m done with this problem. Cheers

7. bump

8. Originally Posted by red and white kop!
bump
Code:
p - q
------
2
is the same thing as (1/2)*(p - q), right?

and 1/2 = 0.5

so, (p - q) / 2 = (1/2)(p - q) = 0.5 (p - q)

There...in nice algebra....you can see that the coordinates given by Soroban are EXACTLY the same points as those given in your text.

9. Originally Posted by Mrspi
Code:
p - q
------
2
is the same thing as (1/2)*(p - q), right? and 1/2 = 0.5
so, (p - q) / 2 = (1/2)(p - q) = 0.5 (p - q)
There...in nice algebra....you can see that the coordinates given by Soroban are EXACTLY the same points as those given in your text
There is a problem with this. Soroban's solution does not agree with the solution given in the text.
The text is correct: $\left( {\frac{{p - q}}{2},\frac{{p + q}}{2}} \right)\;\& \,\left({\frac{{p + q}}{2},\frac{{q - p}}{2}} \right)$
That is not Soroban's solution.

10. Originally Posted by Mrspi
Code:
p - q
------
2
is the same thing as (1/2)*(p - q), right?

and 1/2 = 0.5

so, (p - q) / 2 = (1/2)(p - q) = 0.5 (p - q)

There...in nice algebra....you can see that the coordinates given by Soroban are EXACTLY the same points as those given in your text.
don't be stupid

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