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Set T consists of all points (x,y)(x,y) such that x2+y2=1x2

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Set T consists of all points (x,y)(x,y) such that x2+y2=1x2

by richachampion » Wed Oct 05, 2016 5:49 pm
Set T consists of all points (x,y) such that x²+y²=1. If point (a,b) is selected from set T at random, what is the probability that b>a+1

A. 1/4
B. 1.3
C. 1/2
D. 3/5
E. 2/3
Last edited by richachampion on Fri Oct 14, 2016 3:33 am, edited 1 time in total.
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by richachampion » Wed Oct 05, 2016 5:50 pm
OA: A
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by Matt@VeritasPrep » Fri Oct 14, 2016 2:33 am
x² + y² = 1 is the equation of a circle, centered at the origin, with radius 1.

If you imagine a point inside that circle, the only case in which (the y-coordinate of the point) > 1 + (the x-coordinate of that point) is when the y-coordinate is positive and the x-coordinate is negative. (For instance, consider y = 1/2 and x = -√3/2, or y = √3/2 and x = -1/2.)

Any point in quadrant II will suffice, and no point in any other quadrant will suffice. (If this is confusing, let me know, and I can follow up, but it might make intuitive sense right away.) Since the circle is equally split into the four quadrants, our answer is 1/4.
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by Matt@VeritasPrep » Fri Oct 14, 2016 2:38 am
We could also do this algebraically.

x² + y² = 1

y > x + 1

and since x² ≥ 0, we must have 1 ≥ y², and with it 1 ≥ y.

If y = x + 1, then we have

x² + (x + 1)² = 1

x = 0, y = 1

But since y > x + 1, we must have 0 > x. Since x ≥ -1, we have

1 > y > 0 > x > -1

Now let's think about x² + y² = 1. This has four sets of solutions:

Set 1: y ≥ 0, x ≥ 0
Set 2: y ≥ 0, 0 ≥ x
Set 3: 0 ≥ y, x ≥ 0
Set 4: 0 ≥ y, 0 ≥ x

We learned that y > 0 and 0 > x, so we're in Set 2. The sets are the same size, so Set 2 / All Sets = 1/4.
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by Matt@VeritasPrep » Fri Oct 14, 2016 2:39 am
Whoops! When I said "Any point in quadrant II will suffice" I meant to say "Any point in quadrant II on the circle will suffice" ... but hopefully that was clear anyway, despite my sloppiness!
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