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M19-09

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M19-09 [#permalink]

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If vertices of a triangle have coordinates \((-1, 1)\), \((4, 1)\), and \((x, y)\), what is the area of the triangle?


(1) \(y^2 - 2y - 3 = 0\)

(2) \(x^2 = y^2\)
[Reveal] Spoiler: OA

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(1) \(y^2-2y-3=0\). Solving gives: \(y=3\) or \(y=-1\). Look at the diagram below:
Image
The third vertex is either on line \(y=3\) (blue line) or on line \(y=-1\) (green line). Notice that in either case the height of the triangle is 2 (consider two possible positions of the third vertex shown on the diagram). Now, since the length of the base of the triangle is \(4-(-1)=5\) then the area is \(\frac{1}{2}*base*height=\frac{1}{2}*5*2=5\). Sufficient.

(2) \(x^2=y^2\). This implies that \(|x|=|y|\). Clearly insufficient, consider \(x=y=3\) or \(x=y=5\).


Answer: A
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Re M19-09 [#permalink]

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I think this is a high-quality question and I agree with explanation.
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Re: M19-09 [#permalink]

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New post 23 Jul 2016, 05:47
Alternate method:
x1 x2 x3 x1 --> -1 4 x -1
y1 y2 y3 y1 --> 1 1 y 1

Area of triangle = 0.5 * |(x1y2 - x2y1) + (x2y3 - x3y2) + (x3y1 - x1y3)| --> No need to remember the formula. It is just a simple cross multiplication of the co-ordinates in the above matrix.

Area = 0.5 * |(-1 - 4) + (4y - x) + (x + y)| = 0.5 * |5y - 5|

St1: y^2 - 2y - 3 = 0 --> y = 3 or -1
y = 3 --> Area = 5
y = -1 --> Area = 5
Sufficient

St2: x^2 = y^2 --> |x| = |y|. Clearly insufficient to calculate area.

Answer: A
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Re: M19-09 [#permalink]

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New post 26 Oct 2016, 10:17
What a beautiful question!
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Re: M19-09 [#permalink]

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New post 03 Aug 2017, 06:29
+1 for option A. The area of a triangle is (1/2)*(base)*(height). Base is 5 units long. Height is y co-ordinate minus 1. The area becomes (1/2)*5*(y-1). We need a definite positive value for y. Only statement 1 gives us one. Hence option A.
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Re: M19-09 [#permalink]

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New post 03 Aug 2017, 13:22
Lets use the determinant method to find out the area from the question stem.

Area of triangle = 1/2 |-1(1-y)-1(4-x) +1 (4y-x) = 1/2(y-1 = x-4 =4y-x) = (5y-5)/2 = 5 (y-1)/2
A=5(y-1)/2

Lets look at statement I and II

statement I)
y2-2y-3=0--> y2-3y+y-3 = 0 --> (y-3)(y+1) =0--> y=-1 or y=3
For y=(-1), A= 5/2(-1-1)= -5

For y=3, A=5/2(3-1)= +5

area is always positive , therefore for both cases area is same that is A=5 units

A is sufficient, so cancel out answer B,C,E; only possibilities are A and D

statement II)
x2=y2
x2-y2=0
(x+y)(x-y)=0
either x=-y or x=y
Area is dependent on value of y, since value of y is not defined on basis of a variable x, therefore statement II is not sufficient

Answer= A
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Re: M19-09 [#permalink]

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New post 18 Aug 2017, 06:05
Great question. Did not even think about just needing the height
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Re: M19-09 [#permalink]

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New post 14 Dec 2017, 10:14
Bunuel wrote:
If vertices of a triangle have coordinates \((-1, 1)\), \((4, 1)\), and \((x, y)\), what is the area of the triangle?


(1) \(y^2 - 2y - 3 = 0\)

(2) \(x^2 = y^2\)


Yes, I agree, this question is beautiful! Thanks Bunuel!
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Re: M19-09   [#permalink] 14 Dec 2017, 10:14
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