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In the coordinate plane, a triangle has vertices at (a,0), (b,0), and (x,y). If a>x>b>0>y, which of the following represents the area of that triangle?

A. (ay−by)/2 B. (ab−ay)/2 C. (by−ay)/2 D. (ay+by)/x E. (a−b)/2y

In the coordinate plane, a triangle has vertices at (a,0),(b,0), and ( [#permalink]

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11 Oct 2015, 15:17

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Praveengeol wrote:

Bunuel wrote:

In the coordinate plane, a triangle has vertices at (a,0), (b,0), and (x,y). If a>x>b>0>y, which of the following represents the area of that triangle?

A. (ay−by)/2 B. (ab−ay)/2 C. (by−ay)/2 D. (ay+by)/x E. (a−b)/2y

Kudos for a correct solution.

Answer is A. Please correct if this is wrong.

As this is a forum that is based on discussion, just mentioning the answer without talking about how you approached the question is of no use to anyone. Please do mention your complete solution.

Classic question for assuming numbers for variables.

Let y=-1,b=1,x=3 and a=4

Thus the area of the triangle = (4-1)/2 = 1.5 square units.

Now based on the assumed values, analyse the options

A. (ay−by)/2 = -1.5 . Eliminate. B. (ab−ay)/2 = 4 . Eliminate. C. (by−ay)/2 = 1.5 . Keep. D. (ay+by)/x = -5/3 . Eliminate. E. (a−b)/2y = -1.5 . Eliminate.

Re: In the coordinate plane, a triangle has vertices at (a,0),(b,0), and ( [#permalink]

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11 Oct 2015, 18:59

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in the coordinate plane, a triangle has vertices at (a,0), (b,0), and (x,y). If a>x>b>0>y, which of the following represents the area of that triangle?

It is simplest here to plug in numbers that will satisfy y<0<b<x<a. For example, -1<0<2<3<4

Re: In the coordinate plane, a triangle has vertices at (a,0),(b,0), and ( [#permalink]

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11 Oct 2015, 19:28

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Answer is C

My approach:

The vertex (x, y) lies in the fourth quadrant while the other two vertices are on the x axis. The difference of abscissa gives the base while ordinate value (y) gives the height. Ares = 1/2(base *height)= 1/2(by-ay)

In the coordinate plane, a triangle has vertices at (a,0), (b,0), and (x,y). If a>x>b>0>y, which of the following represents the area of that triangle?

A. (ay−by)/2 B. (ab−ay)/2 C. (by−ay)/2 D. (ay+by)/x E. (a−b)/2y

The area of a triangle is (base)(height)/2, and this problem provides you enough to find a base and corresponding height. Since the y-coordinates of (a,0) and (b,0) are equal, you can tell that that is a horizontal line and serves nicely as your base. That base length, then, is a−b. Then for the height, since the y-coordinates of your base are 0, the value of y will tell you how far the third point ((x,y) is from the x-axis base. However, since you're told that y<0, you'll need to use −y as your height because the triangle cannot have a negative height. That means that (base)(height)/2=(a−b)(−y)/2. The numerator distributes to (−ay+by)/2, and then you can rearrange the addition in the numerator to get to the answer: (by−ay)/2.
_________________

Re: In the coordinate plane, a triangle has vertices at (a,0),(b,0), and ( [#permalink]

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11 Mar 2016, 20:26

Bunuel wrote:

In the coordinate plane, a triangle has vertices at (a,0), (b,0), and (x,y). If a>x>b>0>y, which of the following represents the area of that triangle?

A. (ay−by)/2 B. (ab−ay)/2 C. (by−ay)/2 D. (ay+by)/x E. (a−b)/2y

Kudos for a correct solution.

fastest way - assign values and test.. y<0 -suppose y=-2 b>0 - b=1 x=2 a=3

we have a triangle with base =2, height =2. area thus must be 2.

A: y(a-b)/2 = -2(2)/2 = -2. negative area no. B: a(b-y)/2 = 3(3)/2 = 4.5 - no. C: y(b-a)/2 = -2(-2)/2 = 2. hold D: y(a+b)/x = -2(4)/2 = -4. out E: 2/-4 = -1/2 - negative so out.

Re: In the coordinate plane, a triangle has vertices at (a,0),(b,0), and ( [#permalink]

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24 Apr 2017, 03:47

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This post was BOOKMARKED

Bunuel wrote:

In the coordinate plane, a triangle has vertices at (a,0), (b,0), and (x,y). If a>x>b>0>y, which of the following represents the area of that triangle?

A. (ay−by)/2 B. (ab−ay)/2 C. (by−ay)/2 D. (ay+by)/x E. (a−b)/2y

Kudos for a correct solution.

Hi,

This question can be easily solved if we plot the diagram. Please refer attached diagram.

Base BA = a - b Height XC = -y (because y<0)

Area = \(\frac{1}{2}*(a-b)*(-y) = \frac{by - ay }{2}\)

Thanks.

Attachments

Triangle_Coordinate.jpeg [ 8.23 KiB | Viewed 1176 times ]

Re: In the coordinate plane, a triangle has vertices at (a,0),(b,0), and ( [#permalink]

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16 Jun 2017, 21:58

Bunuel wrote:

In the coordinate plane, a triangle has vertices at (a,0), (b,0), and (x,y). If a>x>b>0>y, which of the following represents the area of that triangle?

A. (ay−by)/2 B. (ab−ay)/2 C. (by−ay)/2 D. (ay+by)/x E. (a−b)/2y

Kudos for a correct solution.

From this question we clearly know that (x, y) will be in the fourth quadrant- so we can just make a simple triangle such as (3,0) (5,0) and (4, -5) - furthermore we can just draw this triangle separately from our coordinate plane to further examine it an avoid confusion- in the scenario listed the height is simply 1/2 * 2 * 5= 10; yet, beyond plugging in numbers in the formulas given by the answer choices shouldn't the answer also be a positive real number? You cannot have a negative area within the context of the GMAT- well perhaps in quantum physics but anyways- the area must be a positive value Bunuel am I right about the value having to be positive?

In the coordinate plane, a triangle has vertices at (a,0), (b,0), and (x,y). If a>x>b>0>y, which of the following represents the area of that triangle?

A. (ay−by)/2 B. (ab−ay)/2 C. (by−ay)/2 D. (ay+by)/x E. (a−b)/2y

Kudos for a correct solution.

From this question we clearly know that (x, y) will be in the fourth quadrant- so we can just make a simple triangle such as (3,0) (5,0) and (4, -5) - furthermore we can just draw this triangle separately from our coordinate plane to further examine it an avoid confusion- in the scenario listed the height is simply 1/2 * 2 * 5= 10; yet, beyond plugging in numbers in the formulas given by the answer choices shouldn't the answer also be a positive real number? You cannot have a negative area within the context of the GMAT- well perhaps in quantum physics but anyways- the area must be a positive value Bunuel am I right about the value having to be positive?

Therefore

"C"

The area obviously must be positive and it turns out to be.

The correct answer is \(\frac{(by−ay)}{2} = -y*\frac{(a-b)}{2} = -negative*positive = positive*positive = positive\).
_________________