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Locus Question 0.0 (1 Viewer)

kev-kun

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This question is just throwing me off. Part ii asks to find the area of the triangle.... No idea on how to go about it. Help is much appreciated!
Untitled picture1.png
 

HSC2014

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Ah, finally worked it out; ending was a little tricky.
Start with what you know about triangles: How do you find the area of one? Which method seems most suitable to your question? Then go for it.
 

kev-kun

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hmmm... either the basic 1/2bh or trig but there's no angle. So the most appropriate is the basic one... hmmm. Base is probably PQ but perpendicular height not sure...
 

HSC2014

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Trust your gut instinct. If you want to excel in higher level math, you have to develop insight - and that is only achievable through experience and fault. As for base, it can be anything in a triangle, depending on perspective. The only question here is which is most efficient for the algebra.
 

HeroicPandas

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There is a reason why they wrote the equation of PQ to be like that instead of
 

SharkeyBoy

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Point O to line PQ is obtained via perp. dist. formula
PQ can be found using distance formula.
Then just 1/2bh
 

HSC2014

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That's okay :) When in doubt, go back to your foundations and use the tools you've learnt in mathematics.
 

RealiseNothing

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A much easier approach:

Drop perpendiculars from the points P and Q to the x-axis and make them the points P' and Q' respectively. So now you have a trapezium PP'Q'Q.

Now Area of OPQ = Area of PP'Q'Q - Area of OPP' - Area of OQQ'







Therefore by subtracting the area of the two triangles from the trapezium we get:

 

Drongoski

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A much easier approach:

Drop perpendiculars from the points P and Q to the x-axis and make them the points P' and Q' respectively. So now you have a trapezium PP'Q'Q.

Now Area of OPQ = Area of PP'Q'Q - Area of OPP' - Area of OQQ'







Therefore by subtracting the area of the two triangles from the trapezium we get:

Nice effort & the diagram too.
 
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