Coordinate

[mkamony]

http://postimage.org/image/3600jxvhg/

Thanks ahead

[jen]

That problem isn't joking around! Here goes.

The point r, t is defined as being in the region below y = 3x - 3 and above y = 1/2x. Therefore, it is in the region y < 3x - 3 and y > 1/2x. Sub in the r and t to get:

t < 3r - 3
t > r/2

You can assemble these, if you like, to get:

r/2 < t < 3r - 3

You can also use the two lines to get the point of intersection (here, setting the y values equal to each other):

3x - 3 = x/2
x = 6/5
plug back in to either equation get y = 3/5

So we could already see that r and t are positive, but now we can see that the point is "up and to the right" from (6/5, 3/5).

On to the choices:

A is plainly true from the previously derived r/2 < t < 3r - 3

B You can prove this to be true by plugging in values, keeping to the constraint that the point is "up and to the right" from (6/5, 3/5).

C True because the point is "up and to the right" from (6/5, 3/5), and 6/5 is already > 1

D The opposite is true! From t < 3r - 3, we get 3r - t > 3.

E We can see that by looking at the picture.

Sincerely,
Jen

p.s. This was remarkably time-consuming, and as such I don't think it's a realistic GRE question.