(P2) 2006 Q.3 (b)

Nehpets

Can someone explain the method to doing this?

I looked at the solutions and don't understand them really.

Steoob

eh... subject?

Carsinian Thau

Draw the diagram and use translations to construct the necessary equations.
Bit tired now so I won't be able to go into the specifics but r will move to p by adding 2 to the x value.
Since pr:pq=1:3. The x value of q can be determined by adding 3 times the difference to p. i.e 2+6=8. -> q=(8,0).
Similarly for q to p the y value decreases by 9. Just subtract one third of this from p to move to r. i.e -9-3=-12. -> r=(0,-12).

Sorry about the lack of clarity.

Carsinian Thau

sorry about my typing but the is supposed to be : p above.
i blame the interweb...

Carsinian Thau

Draw the diagram and use translations to construct the necessary equations.
Bit tired now so I won't be able to go into the specifics but r will move to p by adding 2 to the x value.
Since │pr│:│pq│=1:3. The x value of q can be determined by adding 3 times the difference to p. i.e 2+6=8. -> q=(8,0).
Similarly for q to p the y value decreases by 9. Just subtract one third of this from p to move to r. i.e -9-3=-12. -> r=(0,-12).

Sorry about the lack of clarity.

Much better, anyone know how I get rid of the other two posts?

JC 2K3

Edit, then select to delete em.

genericgoon

FOr this question i actually used the line bisecting formula. Put p=(x,0) and r=(0,y) then using the formula split it into the 1:3 ratio then equating the x bit to the x bit of p and the same for the y part i got x and y and thus the points. I know its not the on in the marking scheme but it works.

Nehpets

Have the points and equations.. I don't understand the moving as in "r will move to p by adding 2 to the x value."

Nehpets

genericgoon wrote:

FOr this question i actually used the line bisecting formula. Put p=(x,0) and r=(0,y) then using the formula split it into the 1:3 ratio then equating the x bit to the x bit of p and the same for the y part i got x and y and thus the points. I know its not the on in the marking scheme but it works.

Thank you! I was trying this earlier but must have done something wrong, just did it there and got it.

Will that always be an option of using that formula? I really don't like that other way.

Argh.... spent all day doing the line.. 5 questions for tomorrow

genericgoon

Yes the only problem with that way is making sure you have the ratio[H:k] and the points in the rite order.

Nehpets

That might have been the problem earlier! Is there anyway to know?

Like say it splits the line segment [ab] in 4:5 then point a is x1y1 and b is x2y2?

genericgoon

Its easy enough if you draw the diagram. In our book it gives a little diagram of the division and thus from it its easy enough to figure it out.

MathsManiac

Also, you can think of it like a weighted mean.

The midpoint is the unweighted mean of the two points. If you want to get closer to point a than point b, you need more of a than b. So if the ratio is, for example, 3:1, and you want to be closer to a than to b, you'll need three times a's co-ordinates and one times b's co-ordinates, (all divided by 4 of course).

It's the same formula in vectors too, as you may have noticed.

Nehpets

Yeah I did vectors after I posted this I still like the formula though