Question

Coordinate geometry (LQ) #2 ( quite a cute one :) )

Answer 1

is this your question paper???

Answer 2

Past paper

Answer 3

Don't worry. The questions I've been asking are NOT homework. I do them as practice

Answer 4

Not even a single clue how to do that.

Answer 5

17a i should be \((x-\frac{1}{2}p)^2+y^2=(\frac{1}{2}p)^2\)

Answer 6

@thomas5267 agreed :)

Answer 7

For b.
Satisfy (a,b) in the equation of the circle. You get:
\[a^2 + b^2 - ap = 0 \]

Answer 8

=> \[a^2 +b^2 = ap\]

=> \[(\sqrt{(a^2 + b^2)}) ^2 = a * p \]

Answer 9

=> \[OS^2 = OR * OP\]

Answer 10

And OQcos(Angle POQ) is OR

Answer 11

Which proves it.

Answer 12

o.o I did it in another way..
OQ cos∠POQ = OR
Join OS and SR
ΔSOP~ΔROS (AAA)
OS OP
-- = --
RO OS

OS^2 = OP x RO = OPxOQ cos∠POQ

Answer 13

OS OP
-- = -- (corr. sides, ~Δs)
RO OS

Answer 14

but good answer! @siddhantsharan
I always don't know how to use previous results :(

Answer 15

HAha.
Your way is pretty tidy and neat. Works good!

Answer 16

lol I missed the proof of similar triangles though. Don't want to type so much :(

Answer 17

Very clever method! @siddhantsharan

Answer 18

Who does, right?
2nd one is pretty straightforward.
Drop perpendicularFrom C to AB ( sOme K)
Then By AA Prove ACK and AEB similar.

Answer 19

@thomas5267 Thanks !

Answer 20

Do you mean (bi)?

Answer 21

For (bi)
Since BC = diameter of the circle
∠BEC =90 (∠s in semi-circle), that is BE⊥AC
So, BE is the height of the ΔABC

Answer 22

I only know that \(\angle CFB=\angle CGA\) . Help Please!

Answer 23

Hmm..If it says using (a), perhaps we can compare them?
C=O
B=P
A=Q
D=R
F=S
perhaps... Would it be useful?

Answer 24

Any known relationship between the diameter of the two circles?

Answer 25

the center of circle in Question 1) are (a+p)/2, 0 right??

Answer 26

i think it is (p/2, 0), not sure

Answer 27

Oh it's p/2 ... I used strange distance formula ... and completely forgot one variable.

Answer 28

for b.) i.)

so for BE to be an altitude, we know it should pass through a vertex of the triangle and that it should be perpendicular to the base. So we let the CA be that base in this case. Now since BE passes already through a vertex of the triangle, we only need to prove that it is perpendicular to our base CA. ANd for that to happen we know that the angle CEB should be a right angle. now applying the thales theorem we know that triangle BCE will be a right triangle where CEB is the right angle, therefore, BE is perpendicular to CA. Since BE is perpendicular to CA and passes through a vertex of the triangle which is B, then it is an altitude of the triangle ABC

Answer 29

@anonymoustwo44 I think using ∠s in semi-circle is a pretty good way to prove it :)

Answer 30

But thanks for providing another solution :)

Answer 31

Theorem: An angle inscribed in a Semi-circle is a right angle.

Answer 32

The only part left is (bii), any ideas?

Answer 33

@Callisto, ah I didn't know we had that :)) I only knew thales theorem thanks for making me know this thing. Sorry for making the proof long, I was just afraid it wouldn't satisfy your level of rigourosity

Answer 34

Wait, notice that 17b(ii) says that JUSTIFY our answer not PROOF our answer. Then I say that \(CF>CG\) as \(CF^2+FB^2=CB^2>CA^2=CG^2+GA^2\) . Hehe!

Answer 35

@anonymoustwo44 The proof is good. And I've never thought of it. Thanks for bringing it up here!!! :)

Answer 36

@thomas5267 Hmmm.. good point. (for justify and prove) But... can you explain a little?! I know you're using Pyth. Thm.. and through comparing the hypotenuse, you want to get something. But I'm a little confused!

Answer 37

\[CB^{2} > AC^{2}\] then \[CF > CG\]
hmmm i don't think so hehe

Answer 38

That's because we have many combinations for the unknowns. So, it's difficult to justify it in that way

Answer 39

Just JUSTIFYING not proofing. Just playing with words...

Answer 40

Yup~ but you need to be reasonable in justifying sth, I think :S

Answer 41

But as the picture is not to scale, the diameter of both circle can be anything. What can you proof then?

Answer 42

lol, if I could do it now, I would close the question ...