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M19 Q9

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Senior Manager
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M19 Q9 [#permalink]

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New post 12 Nov 2008, 21:38
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. \(x^2 = y^2\)

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Re: M19 Q9 [#permalink]

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New post 12 Nov 2008, 22:21
It should be A.

From stmt1: y = 3 or -1. In either case, base of triangle is 3 and height is 2 and area can be calculated..

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Re: M19 Q9 [#permalink]

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New post 12 Nov 2008, 22:29
Butwhat if x is -4 and y is -1 (-4,-1)
would the height of the triangle still be 2?
I apologize if this may seem a stupid question but when I drew this out it wasnt clear hence the confusion with this question.
Thank You

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New post 12 Nov 2008, 22:39
ventivish wrote:
Butwhat if x is -4 and y is -1 (-4,-1)
would the height of the triangle still be 2?
I apologize if this may seem a stupid question but when I drew this out it wasnt clear hence the confusion with this question.
Thank You


Yes.

Since the line joining (1,1,) and (4,1) is parallel to x-axis, this can be treated as the base.

For the third vertex, it is the y co-ordinate that will decide the height of the triangle. If the y co-ordiate of 3, height of triangle is 2. If y-co-ordiate is -1, then also, height of triangle is 2.

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Re: M19 Q9 [#permalink]

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New post 12 Nov 2008, 23:03
ventivish wrote:
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. \(x^2 = y^2\)


nice job scthakur.

thank ventivish for bringing up thing question.
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Re: M19 Q9 [#permalink]

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New post 18 Nov 2008, 04:33
Thanks for replying to this!!!

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Re: M19 Q9   [#permalink] 18 Nov 2008, 04:33
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