Hi. I have a question that has stumped me. Any help would be appreciated.
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.
I seem to be missing an ingredient to the answer. I can find the mid-point and the length of OB. I can also find the gradient of OB and of AC. But none of that really helps unless I can find a relationship between the points.
The answer is [1/2(p-q), 1/2(p q)], [1/2(p q), 1/2(q-p)]. Any guidance would be appreciated.

Hi. I have a question that has stumped me. Any help would be appreciated.
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.
I seem to be missing an ingredient to the answer. I can find the mid-point and the length of OB. I can also find the gradient of OB and of AC. But none of that really helps unless I can find a relationship between the points.
The answer is [1/2(p-q), 1/2(p q)], [1/2(p q), 1/2(q-p)]. Any guidance would be appreciated.

I had this question come up in AS Level mathematics and I'm stumped.
ABC is a triangle. D is the mid-point of BC, E is the mid-point of AC, F is th amid-point of AB and G is the mid-point of EF. Express the displacement vectors AD and AG in terms of a, b, and c. The answer for AD is 1/2(b c -2a) and for AG is 1/4(b c -2a). I assume that the answer for vector AD can be written as 1/2(b c) - a. I understand where the 1/2(b c) came from, but do not understand where the -a came from. If I use a position vector, I assume it should make a parallelogram OABC but I still do not understand why -a.
Any help would be appreciated. Thanks. Adrian.

To recognize another year's usage of capital assets?

Dr: amortization expense
CR: accumulated ammortization
Does this look like the proper entryto record ammortization expense at year end to recognize another year's usage of capital assets, in the generation of revenue?

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Originally Posted by AdrianCavinder

Hi. I have a question that has stumped me. Any help would be appreciated.
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.
I seem to be missing an ingredient to the answer. I can find the mid-point and the length of OB. I can also find the gradient of OB and of AC. But none of that really helps unless I can find a relationship between the points.
The answer is [1/2(p-q), 1/2(p q)], [1/2(p q), 1/2(q-p)]. Any guidance would be appreciated.

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You can solve this by finding the angle of OB, which is Arctan(q,p), so the angle of OA is Arctab{q/p)-pi/4. The coordinates of point A are then:

(L sin(arctan(q/p - pi/4), L cos(arctan(q/p-pi/4))

Where L - length of a side of the square, or sqrt((p^2+q^2)/2). Using sin(a+b) = sin(a)cos(b) + cos(a)sin(b) you get for eth firsrt coordinate of point A:



You can use this same technique to find the 2nd coordinate of A. Then for point C change the -pi/4 parts to +pi/4 and repeat the process.

Dear Uber, Thanks again. I'm sure the method is correct, the only problem is that this question is in a revision paper for UK AS level mathematics (equivalent to US 11th grade). As such, your explanation is too complex as much of what you describe hasn't been covered within the syllabus. I assume there must be a simpler explanation somehow. Anyway, I certainly appreciate your answer and taking the time to respond.
Btw, I only repeated the question because I got an e-mail message to 'bump' the question and I had no idea what that meant: I assumed it meant change the wording around a bit and re-submit. If you have any input on what 'bumping' is I'd appreciate it. Thanks.

It's interesting that trigonometry functions haven't been taught in your course yet - I learned sines, cosines, etc in 10th grade, though in many schools it's not covered until beginning of 11th.

Another way to do his, though much messier, is with algebra. You can create two equations in two unknowns as follows.

Call the point of one of the corners (a,b). The line from the origin through point (a,b) has the slope (b/a), and its equation is:

y = (b/a)x

The line that goes from (a,b) to the point (p,q) is perpendicular to this line, and hence has slope -a/b (recall that perpendicular lines have slopes that are negative inverses of each other). The equation of a line with that slope that passes through (p,q) is:

y-q = (-a/b)(x-p)

Since the point (a,b) lies on that line, you have:

b-q = (-a/b)(a-p)

Play with this a bit and you get: a^2 + b^2 = ap+bq. This is your first equation.

The other thing we know is that for square the length of the diagonal is sqrt(2) times the length of one of the sides. Hence:

p^2 + q^2 = 2(a^2 + b^2)

Now you have two equations in two unknowns (a and b). It takes a bit of work to solve these, but you do end up with the values for a and b that give the correct result.