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Re: M16-15 [#permalink]
I'm confused how we know that Only the angle at vertex (0,A)(0,A) can be right.
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Re: M16-15 [#permalink]
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joannaecohen wrote:
I'm confused how we know that Only the angle at vertex (0,A)(0,A) can be right.


Plot the points. Other vertices to be right the third vertex should be either on (-1, y) or at (4, y).
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Re: M16-15 [#permalink]
I think another way to approach this problem is by considering the following.

If you draw out the triangle you notice that (−1,0), (4,0) Lie on the X axis with a distance of 5. The third point lies at some number "a" the Y axis.

(1) a>3

Not Sufficient because if a = 4 then the area is 4*5*1/2 = 10 which is less than 15. Now if a = 10 then the area is 5*10*1/2 = 25 which is greater than 15.

(2) The Triangle is Right.

Sufficient. If you quickly plot the points, you will realize that in order for the triangle to be right the side that measures 5 on the X axis has to be the hypotenuse.

Using Pythagorean's Theorem substituting 5 for C you can see that a^2 + b^2 = 5^2 This tells us that the Base and Height must be less than 5. Therefore no combination of a*b*1/2 will give you a number greater than 15.

So the Answer Choice Must be B.
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Re: M16-15 [#permalink]
I interpreted question stem as: is 1/2*5*|A|>15?
If A>0 => 5A/2>15 or is A>6?
If A<0 => -5A/2>15 or is A<-6

(1) A<3. Doesn't rule out A=2,1,-1,-2.. etc, so insufficient

(2) Right triangle. Neither of the vertices on AB are at a perpendicular to vertex C (AC and BC aren't vertical), therefore the right angle must be at ACB. Thus, per Pythagoras, 5^2 = AC^2 + BC^2, meaning that both AC & BC < 5 and by extension, Area = 1/2*AC*BC < 15. Sufficient.
(p.s. the various methods on here for assessing statement (2) are all great)
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Re: M16-15 [#permalink]
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M16-15 [#permalink]
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