Find P(X,Y), given P(x,y).

Question

The line y=x+1 forms the Y axis of another set of rectangular XY axes with origin at (1,2) in the xy plane (i.e., the point (1,2) in the xy plane is the point (0,0) in the XY plane). Find P(X,Y) for the general point P(x,y) in the xy plane (i.e., find X and Y in terms of x and y).

Comments

It seems hard to find the answer to that. I don't know Rod. Sorry. :(

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When you've done graphs and can use geometry you may then be able to answer this question, which just needs thought, effort and time. A picture helps to answer this question :-)

Answer 1

So we have P(X>Y)=P(X−Y>0)=P((X,Y)∈C)P(X>Y)=P(X−Y>0)=P((X,Y)∈C)

Where C={(x,y):x−y>0}C={(x,y):x−y>0}

But we know how to find this, as this is the same as simply evaluating f(x,y) and integrating this over the region C.

P(X>Y)=∬Cf(x,y)dxdyP(X>Y)=∬Cf(x,y)dxdy .

Comments

I can see you worked hard to find an answer, but it is in the wrong area of math. The answer only involves graphs and geometry, and does not use calculus (so does not use integration). When you draw a picture it helps to see the way towards a solution. You will learn how to do this  one day when you are a bit older. I didn't know how to do this when I was your age. And there was no internet then!