Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 12 Feb 2012
Posts: 135

Triangle T has vertices of (2,6), (6,2) and (0,k) where k [#permalink]
Show Tags
12 Jul 2012, 17:29
2
This post received KUDOS
5
This post was BOOKMARKED
Question Stats:
49% (03:03) correct
51% (01:51) wrong based on 104 sessions
HideShow timer Statistics
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(y3y2) + x2(y1y3) +x3(y2y1] 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(sa)(sb)(sc)) where s = (a+b+c)/2 Help!! Thank you!
Official Answer and Stats are available only to registered users. Register/ Login.



Intern
Joined: 22 Jan 2012
Posts: 35

Re: The Area of a Triangle with only 3 Vertices Known [#permalink]
Show Tags
12 Jul 2012, 18:35
1
This post received KUDOS
Use the formula:
1/2(x1x3)(y2y1) (x1x2) (y3y1)
= 4k20
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.



Intern
Joined: 25 Jun 2012
Posts: 7

Re: The Area of a Triangle with only 3 Vertices Known [#permalink]
Show Tags
13 Jul 2012, 19:11
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 454590 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 = 3k.
Triangle T is the sum of the 3 considered triangles and has area T = 20 + 4k = 20  4k, which coinsides with the formula the other poster used.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7446
Location: Pune, India

Re: The Area of a Triangle with only 3 Vertices Known [#permalink]
Show Tags
15 Jul 2012, 23:59
2
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
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(y3y2) + x2(y1y3) +x3(y2y1]
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(sa)(sb)(sc)) 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)
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16014

Re: Triangle T has vertices of (2,6), (6,2) and (0,k) where k [#permalink]
Show Tags
02 Feb 2014, 07:44
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources



Manager
Joined: 17 Mar 2014
Posts: 70

Re: Triangle T has vertices of (2,6), (6,2) and (0,k) where k [#permalink]
Show Tags
09 May 2014, 04:09
Area=1/2 * [x1(y3y2) + x2(y1y3) +x3(y2y1] as posted by alphabeta1234
Area = 1/2(x1x3)(y2y1) (x1x2) (y3y1) 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



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7446
Location: Pune, India

Re: Triangle T has vertices of (2,6), (6,2) and (0,k) where k [#permalink]
Show Tags
12 May 2014, 21:08
qlx wrote: Area=1/2 * [x1(y3y2) + x2(y1y3) +x3(y2y1] as posted by alphabeta1234
Area = 1/2(x1x3)(y2y1) (x1x2) (y3y1) 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(x1x3)(y2y1) (x1x2) (y3y1) Area = 1/2x1(y2y1)  x3(y2  y1)  x1(y3y1) + x2(y3  y1) Area = 1/2x1(y2  y3) + x2(y3  y1)  x3(y2  y1) Area = 1/2x1(y2  y3) + x2(y3  y1) + x3(y1  y2) This is the third formula.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Manager
Joined: 17 Mar 2014
Posts: 70

Re: Triangle T has vertices of (2,6), (6,2) and (0,k) where k [#permalink]
Show Tags
13 May 2014, 02:00
VeritasPrepKarishma wrote: qlx wrote: Area=1/2 * [x1(y3y2) + x2(y1y3) +x3(y2y1] as posted by alphabeta1234
Area = 1/2(x1x3)(y2y1) (x1x2) (y3y1) 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(x1x3)(y2y1) (x1x2) (y3y1) Area = 1/2x1(y2y1)  x3(y2  y1)  x1(y3y1) + x2(y3  y1) Area = 1/2x1(y2  y3) + x2(y3  y1)  x3(y2  y1) Area = 1/2x1(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 1/2x1(y2 y3)+x2(y3y1)+x3(y1y2) 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.



Manager
Joined: 14 Jul 2014
Posts: 191
Location: United States
GMAT 1: 600 Q48 V27 GMAT 2: 720 Q50 V37
GPA: 3.2

Re: Triangle T has vertices of (2,6), (6,2) and (0,k) where k [#permalink]
Show Tags
22 Dec 2015, 20:47
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.



Manager
Joined: 20 Mar 2016
Posts: 179
Location: Estonia
Concentration: Strategy, General Management

Re: Triangle T has vertices of (2,6), (6,2) and (0,k) where k [#permalink]
Show Tags
19 May 2016, 09:53
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 454590 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 = 3k.
Triangle T is the sum of the 3 considered triangles and has area T = 20 + 4k = 20  4k, which coinsides with the formula the other poster used. This seems Wrong at some point. Area is 20+ 4k. where are k and 3k 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 204K Clear. Now to make this a multiple of 3 or 5. the area 4(5K) must be a multiple of 3 or 5. so (5K) must be multiple of 3 or 5. correct me if i am wrong.



Director
Joined: 24 Nov 2015
Posts: 561
Location: United States (LA)

Re: Triangle T has vertices of (2,6), (6,2) and (0,k) where k [#permalink]
Show Tags
19 May 2016, 13:40
Formula stated as 1/2 ( x1(y2  y3) + x2(y3  y1) + x3(y1  y2)) or 1/2(x1x3)(y2y1) (x1x2) (y3y1) 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



Manager
Joined: 20 Mar 2016
Posts: 179
Location: Estonia
Concentration: Strategy, General Management

Re: Triangle T has vertices of (2,6), (6,2) and (0,k) where k [#permalink]
Show Tags
19 May 2016, 19:12
rhine29388 wrote: Formula stated as 1/2 ( x1(y2  y3) + x2(y3  y1) + x3(y1  y2)) or 1/2(x1x3)(y2y1) (x1x2) (y3y1) 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 xy 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 yaxis = 2 (0,k), (6,2) and (0,0) = base is k height is distance of (6,2) from yaxis = 6




Re: Triangle T has vertices of (2,6), (6,2) and (0,k) where k
[#permalink]
19 May 2016, 19:12








Similar topics 
Author 
Replies 
Last post 
Similar Topics:




In the equation x^2 + 5x + k = 0, k is a constant and x is a variable.

Bunuel 
1 
11 Nov 2016, 03:35 

1


If km does not equal 0, is k/m < 0?

Bunuel 
2 
12 Oct 2016, 04:08 



If t > 0, is k < 8

ritumaheshwari02 
3 
27 Nov 2012, 03:40 

7


If n and k are integers and (2)n^5 > 0, is k^37 < 0?

pratikbais 
7 
26 Jun 2015, 02:20 

107


If vertices of a triangle have coordinates (2,2), (3,2) and

noboru 
37 
20 Apr 2017, 10:04 



