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Triangle T has vertices of (-2,6), (6,2) and (0,k) where k

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Triangle T has vertices of (-2,6), (6,2) and (0,k) where k [#permalink] New post 12 Jul 2012, 16:29
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Triangle T has vertices of (-2,6), (6,2) and (0,k) where k is negative. Is k an integer?

(1) The area of T in square units is a multiple of 3
(2) The area of T in square units is a multiple of 5


Can this formula be used to find the area of any triangle with three vertices? Or must the triangle be a right angle for this equation to be applicable. Does anyone know? I found it used by another user on Gmatclub, not sure if its correct.

Area=1/2 * [x1(y3-y2) + x2(y1-y3) +x3(y2-y1]

The other method I could think of was posted by Bunuel, Hero's formula. But I tried using it, the equation was too sloppy for this answer.

Once you have 3 sides (a,b,c), then the area is calculated using

A = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2



Help!! Thank you!
[Reveal] Spoiler: OA
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Re: The Area of a Triangle with only 3 Vertices Known [#permalink] New post 12 Jul 2012, 17:35
Use the formula:

1/2|(x1-x3)(y2-y1)- (x1-x2) (y3-y1)|

= |4k-20|

Since K is negative

1) Area is a multiple of 3 , so k=-1,-2,-3
yields area= 24,28,32 ....
and 24,36 are multiples of 3
Not sufficient as K can be integer as well as real

2)Area is a multiple of 5 , so k=-1,-2,-3
yields area= 25,30,35 ....
and 25,30,35 are multiples of 5
Not sufficient as K can be integer as well as real

1&2)Area is a multiple of 3,5
for k = -10
Area= 60
Not sufficient as K can be integer as well as real

Hence , the answer is E.
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Re: The Area of a Triangle with only 3 Vertices Known [#permalink] New post 13 Jul 2012, 18:11
The formula seems correct but GMAT doesn't rely on such criptic formulas.

I would calculate the area by first considering the triangle with vertices (-2, 6), (6,2) and the origin (0, 0). As usual in GMAT, it turns out a special triangle. It is an isosceles right triangle (in other words 45-45-90 triangle) with legs = root(6^2+2^2) = root(40). It's area is leg*leg/2 = 20.

Then I would consider the triangle with vertices (-2, 6), (k, 0) and the origin (0 , 0). It's base=|k| and height to that base = 2. So it's area is |k|*2/2 = |k|.

Last, I would consider the triangle with vertices (2, 6), (k, 0) and the origin (0 , 0). It's base=|k| and height to that base = 6. So it's area is |k|*6/2 = 3|k|.

Triangle T is the sum of the 3 considered triangles and has area T = 20 + 4|k| = 20 - 4k, which coinsides with the formula the other poster used.
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Re: The Area of a Triangle with only 3 Vertices Known [#permalink] New post 15 Jul 2012, 22:59
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alphabeta1234 wrote:
Triangle T has vertices of (-2,6), (6,2) and (0,k) where k is negative. Is k an integer?

(1) The area of T in square units is a multiple of 3
(2) The area of T in square units is a multiple of 5


Can this formula be used to find the area of any triangle with three vertices? Or must the triangle be a right angle for this equation to be applicable. Does anyone know? I found it used by another user on Gmatclub, not sure if its correct.

Area=1/2 * [x1(y3-y2) + x2(y1-y3) +x3(y2-y1]

The other method I could think of was posted by Bunuel, Hero's formula. But I tried using it, the equation was too sloppy for this answer.

Once you have 3 sides (a,b,c), then the area is calculated using

A = sqrt(s(s-a)(s-b)(s-c)) where s = (a+b+c)/2



Help!! Thank you!


This question is painful without the formula but GMAT doesn't require you to know anything other than the basic formulas. So I wouldn't expect anything like this in actual GMAT.

Using the formula, you can do it quite easily.
Area = |1/2 ( x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))|

When you put in the values of the coordinates, you get Area = |4k - 20|

Say, if area = 15 (a multiple of 3 and 5), k = 35/4 (not an integer)
If area = 60 (a multiple of 3 and 5), k = 20 (an integer)

So k can be an integer or non integer. Answer (E)
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Re: Triangle T has vertices of (-2,6), (6,2) and (0,k) where k [#permalink] New post 02 Feb 2014, 06:44
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Re: Triangle T has vertices of (-2,6), (6,2) and (0,k) where k [#permalink] New post 09 May 2014, 03:09
Area=1/2 * [x1(y3-y2) + x2(y1-y3) +x3(y2-y1] as posted by alphabeta1234

Area = 1/2|(x1-x3)(y2-y1)- (x1-x2) (y3-y1)| as posted by raingary

Area = |1/2 ( x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))|as posted by VeritasPrepKarishma

Seeing the same formula represented in 3 different ways has confused me a little
Can someone explain the formula in the most basic form, giving the necessary fundamentals.

And show how we are arriving to the three different forms as shown above.
Thank you
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Re: Triangle T has vertices of (-2,6), (6,2) and (0,k) where k [#permalink] New post 12 May 2014, 20:08
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qlx wrote:
Area=1/2 * [x1(y3-y2) + x2(y1-y3) +x3(y2-y1] as posted by alphabeta1234

Area = 1/2|(x1-x3)(y2-y1)- (x1-x2) (y3-y1)| as posted by raingary

Area = |1/2 ( x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))|as posted by VeritasPrepKarishma

Seeing the same formula represented in 3 different ways has confused me a little
Can someone explain the formula in the most basic form, giving the necessary fundamentals.

And show how we are arriving to the three different forms as shown above.
Thank you


Note that the first and the third formulas are the same. Since we are looking for area which cannot be negative, it is an absolute value. Note | | around the area.
If you multiply the first formula by -1, you get the third formula.

Also, the second formula is the same as the third formula too. Just open the brackets to see that.

Area = 1/2|(x1-x3)(y2-y1)- (x1-x2) (y3-y1)|
Area = 1/2|x1(y2-y1) - x3(y2 - y1) - x1(y3-y1) + x2(y3 - y1)|
Area = 1/2|x1(y2 - y3) + x2(y3 - y1) - x3(y2 - y1)|
Area = 1/2|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
This is the third formula.
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Re: Triangle T has vertices of (-2,6), (6,2) and (0,k) where k [#permalink] New post 13 May 2014, 01:00
VeritasPrepKarishma wrote:
qlx wrote:
Area=1/2 * [x1(y3-y2) + x2(y1-y3) +x3(y2-y1] as posted by alphabeta1234

Area = 1/2|(x1-x3)(y2-y1)- (x1-x2) (y3-y1)| as posted by raingary

Area = |1/2 ( x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))|as posted by VeritasPrepKarishma

Seeing the same formula represented in 3 different ways has confused me a little
Can someone explain the formula in the most basic form, giving the necessary fundamentals.

And show how we are arriving to the three different forms as shown above.
Thank you


Note that the first and the third formulas are the same. Since we are looking for area which cannot be negative, it is an absolute value. Note | | around the area.
If you multiply the first formula by -1, you get the third formula.

Also, the second formula is the same as the third formula too. Just open the brackets to see that.

Area = 1/2|(x1-x3)(y2-y1)- (x1-x2) (y3-y1)|
Area = 1/2|x1(y2-y1) - x3(y2 - y1) - x1(y3-y1) + x2(y3 - y1)|
Area = 1/2|x1(y2 - y3) + x2(y3 - y1) - x3(y2 - y1)|
Area = 1/2|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
This is the third formula.



Thank you did some google searching and found out that the area of a triangle, given the coordinates of its 3 vertices is given by 1/2|x1(y2- y3)+x2(y3-y1)+x3(y1-y2)|
where x1y1, x2y2, x3y3 are the coordinates of the 3 vertices.
This was what I wanted someone to tell me as I did not know that.
This was a great new formula that I learnt.Thank you.
Re: Triangle T has vertices of (-2,6), (6,2) and (0,k) where k   [#permalink] 13 May 2014, 01:00
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