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Triangle T has vertices of (-2,6), (6,2) and (0,k) where k [#permalink]

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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

Re: The Area of a Triangle with only 3 Vertices Known [#permalink]

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12 Jul 2012, 17:35

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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

Re: The Area of a Triangle with only 3 Vertices Known [#permalink]

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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.

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)
_________________

Re: Triangle T has vertices of (-2,6), (6,2) and (0,k) where k [#permalink]

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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]

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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

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.
_________________

Re: Triangle T has vertices of (-2,6), (6,2) and (0,k) where k [#permalink]

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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]

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22 Dec 2015, 19:47

I'm not sure if this is the right way - I just found the height of the triangle (5 + k) and base ( 8) and got the area as (20 +4k). Then, I just tried to solve for k and eventually got E as the answer.

Re: Triangle T has vertices of (-2,6), (6,2) and (0,k) where k [#permalink]

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19 May 2016, 08:53

tutorphd wrote:

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.

This seems Wrong at some point. Area is 20+ 4|k|. where are |k| and 3|k| is area of Triangle 2 and 3. Total area would be addition of these three triangles as k is below axis. so the total area must be 20-4K Clear.

Now to make this a multiple of 3 or 5. the area 4(5-K) must be a multiple of 3 or 5. so (5-K) must be multiple of 3 or 5.

Re: Triangle T has vertices of (-2,6), (6,2) and (0,k) where k [#permalink]

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19 May 2016, 12:40

Formula stated as |1/2 ( x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))| or 1/2|(x1-x3)(y2-y1)- (x1-x2) (y3-y1)| is actually derived from determinants which i am pretty sure is out of scope of GMAT exam Is there any other method to solve the given problem without using the above formula? Expert help needed

Re: Triangle T has vertices of (-2,6), (6,2) and (0,k) where k [#permalink]

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19 May 2016, 18:12

rhine29388 wrote:

Formula stated as |1/2 ( x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))| or 1/2|(x1-x3)(y2-y1)- (x1-x2) (y3-y1)| is actually derived from determinants which i am pretty sure is out of scope of GMAT exam Is there any other method to solve the given problem without using the above formula? Expert help needed

yes there is. it might not be the best one: Draw the points on x-y plane. now you can split the triangle in three (-2,6), (6,2) and (0,0) = isosceles right triangle with base and height = distance of both the points from (0,0) = sq root (40) (-2,6), (0,k) and (0,0) = base is k height is distance of (-2,6) from y-axis = 2 (0,k), (6,2) and (0,0) = base is k height is distance of (6,2) from y-axis = 6

gmatclubot

Re: Triangle T has vertices of (-2,6), (6,2) and (0,k) where k
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19 May 2016, 18:12

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