M19 Q9 : Retired Discussions [Locked]
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# M19 Q9

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Senior Manager
Joined: 20 Feb 2008
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12 Nov 2008, 20:38
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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12 Nov 2008, 21:21
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..
Senior Manager
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12 Nov 2008, 21:29
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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12 Nov 2008, 21:39
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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12 Nov 2008, 22:03
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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Senior Manager
Joined: 20 Feb 2008
Posts: 296
Location: Bangalore, India
Schools: R1:Cornell, Yale, NYU. R2: Haas, MIT, Ross
Followers: 4

Kudos [?]: 45 [0], given: 0

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18 Nov 2008, 03:33
Re: M19 Q9   [#permalink] 18 Nov 2008, 03:33
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# M19 Q9

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