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Coordinate Geometry

Here is question which I'm stuck on:

A triangle has vertices P(-2,2), Q(q,0) and R(5,3).
The side PQ is twice as long as side QR.
Find the possible values of q.

How do I answer this question??:confused::confused:
Reply 1
Avatar for Oromis263
Oromis263
2
Original post by krisshP
Here is question which I'm stuck on:

A triangle has vertices P(-2,2), Q(q,0) and R(5,3).
The side PQ is twice as long as side QR.
Find the possible values of q.

How do I answer this question??:confused::confused:


Try it in terms of vectors. :smile: I can post a start if you want it.
Reply 2
Avatar for Notnek
Notnek
21
PQ=2QR

What is the length of the line PQ? What is the length of the line QR?

Post some working.
Reply 3
Avatar for krisshP
krisshP
OP
14
Original post by notnek
PQ=2QR

What is the length of the line PQ? What is the length of the line QR?

Post some working.


(q−5)2+(0−3)2=12PQ\sqrt{(q-5)^2+(0-3)^2}=\frac{1}{2}PQ

Am I heading the right way?? Shall I continue from here?
Reply 4
Avatar for Notnek
Notnek
21
Original post by krisshP
(q−5)2+(0−3)2=12PQ\sqrt{(q-5)^2+(0-3)^2}=\frac{1}{2}PQ

Am I heading the right way?? Shall I continue from here?

That's correct so far. Now write down the length of PQ in terms of q.
Reply 5
Avatar for krisshP
krisshP
OP
14
(q−5)2+(0−3)2=12PQ\sqrt{(q-5)^2+(0-3)^2}=\frac{1}{2}PQ

(q−5)2+(−3)2=12PQ\sqrt{(q-5)^2+(-3)^2}=\frac{1}{2}PQ

q2−5q−5q+25+9=12PQ\sqrt{q^2-5q-5q+25+9}=\frac{1}{2}PQ

q2−10q+34=12PQ\sqrt{q^2-10q+34}=\frac{1}{2}PQ

Now what???:confused:
(edited 11 years ago)
Reply 6
Avatar for Notnek
Notnek
21
Original post by krisshP
(q−5)2+(0−3)2=12PQ\sqrt{(q-5)^2+(0-3)^2}=\frac{1}{2}PQ

(q−5)2+(−3)2=12PQ\sqrt{(q-5)^2+(-3)^2}=\frac{1}{2}PQ

q2−5q−5q+25+9=12PQ\sqrt{q^2-5q-5q+25+9}=\frac{1}{2}PQ

q2−10q+34=12PQ\sqrt{q^2-10q+34}=\frac{1}{2}PQ

Now what???:confused:

What is PQ? Use the same method as you used to find QR.
Reply 7
Avatar for krisshP
krisshP
OP
14
ok notnek I will.

q2−10q+34=12×(−2−q)2+(2−0)2\sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{(-2-q)^2+(2-0)^2}

q2−10q+34=12×(−2−q)2+(2)2\sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{(-2-q)^2+(2)^2}

q2−10q+34=12×(−2−q)2+4\sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{(-2-q)^2+4}

q2−10q+34=12×q2+2q+2q+4+4\sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{q^2+2q+2q+4+4}

q2−10q+34=12×q2+4q+8\sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{q^2+4q+8}

Now what do I do??
(edited 11 years ago)
Reply 8
Avatar for Notnek
Notnek
21
Original post by krisshP
ok notnek I will.

q2−10q+34=12×(−2−q)2+(2−0)2\sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{(-2-q)^2+(2-0)^2}

q2−10q+34=12×(−2−q)2+(2)2\sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{(-2-q)^2+(2)^2}

q2−10q+34=12×(−2−q)2+4\sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{(-2-q)^2+4}

q2−10q+34=12×q2+2q+2q+4+4\sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{q^2+2q+2q+4+4}

q2−10q+34=12×q2+4q+8\sqrt{q^2-10q+34}=\frac{1}{2}\times\sqrt{q^2+4q+8}

Now what do I do??

Now square both sides so you get:

Unparseable latex formula:

\displaystyle q^2-10q+34 = \frac{1}{4}\left(q^2+4q+8)



Can you simplify and solve this?
Reply 9
Avatar for krisshP
krisshP
OP
14
Original post by notnek
Now square both sides so you get:

Unparseable latex formula:

\displaystyle q^2-10q+34 = \frac{1}{4}\left(q^2+4q+8)



Can you simplify and solve this?


I'll try to do the rest.

Thanks a lot for your help :smile::smile::smile::smile::smile::smile:
Reply 10
Avatar for krisshP
krisshP
OP
14
Original post by notnek
Now square both sides so you get:

Unparseable latex formula:

\displaystyle q^2-10q+34 = \frac{1}{4}\left(q^2+4q+8)



Can you simplify and solve this?


I end up with the following:

−0.75q2+11q−32=0-0.75q^2+11q-32=0

Is this right?? Can I still solve it??
(edited 11 years ago)
Reply 11
Avatar for Notnek
Notnek
21
Original post by krisshP
I end up with the following:

−0.75q2+11q−32=0-0.75q^2+11q-32=0

Is this right?? Can I still solve it??

That's correct. You can tidy the equation up and make it easier to solve by multiplying it by -4.
Reply 12
Avatar for krisshP
krisshP
OP
14
Original post by notnek
That's correct. You can tidy the equation up and make it easier to solve by multiplying it by -4.


I get the following:
3q2−44q+128=03q^2-44q+128=0

Is it ok for me to divide both sides by 3 and use the completing the square method?
Reply 13
Avatar for krisshP
krisshP
OP
14
I got the answer:

q=4 and q=32/3

Thanks a lot Notnek for your help :smile:
Reply 14
Avatar for Notnek
Notnek
21
Original post by krisshP
I get the following:
3q2−44q+128=03q^2-44q+128=0

Is it ok for me to divide both sides by 3 and use the completing the square method?

You can use that method here but I've always been more of a fan of factorising:

3q2−44q+128=(3q−...)(q−...)3q^2-44q+128 = (3q-...)(q-...)

If you really aren't able to factorise then completing the square is OK to use.
Reply 15
Avatar for Oromis263
Oromis263
2
Original post by krisshP
I get the following:
3q2−44q+128=03q^2-44q+128=0

Is it ok for me to divide both sides by 3 and use the completing the square method?


That, or using the quadratic formula. (or factorising, of course, hadn't checked it to see if it did) :smile:

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