In the coordinate plane, a triangle has vertices at (a,0),(b,0), and (

Question

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

Bunuel · Accepted Answer

VERITAS PREP OFFICIAL SOLUTION:

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.

ENGRTOMBA2018 · Answer

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.

C is the correct answer.

peachfuzz · Answer

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

Only C satisfies the area of a triangle.

Answer:
C. (by−ay)/2

longfellow · Answer

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)

riteshgmat · Answer

Area of the triangle with vertices (x1, y1) , (x2, y2) , (x3, y3) = \frac{1}{2}

We are given vertices as (a,0), (b,0), and (x,y)

So area of triangle is =  \frac{1}{2}

thus we have area of triangle =  \frac{(by−ay)}{2}

mvictor · Answer

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.

only C works.

ganand · Answer

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.

Nunuboy1994 · Answer

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"

Bunuel · Answer

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 .

shinrai15 · Answer

My approach!

@Bunnel Please let me know if method is correct:

Vertices given (a,0), (b,0), (x,y)

I found the slope between all the points

Slope between (a, 0) and (b,0) = 0/b-a
Slope between (b,0) and (x,y) = y/x-b

Equate slopes =(b-a)(y) =by -ay
Area = by-ay/2

Thanks in advance!
S

baraa900 · Answer

how did you get the base? a-b

Fdambro294 · Answer

1st Vertex: (a,0) will be on the X Axis and will be the furthest X Coordinate to the Right (say 10) ———> (10 , 0)

2nd Vertex: (x, y) will have a (+)Positive X Coordinate and a (-)Negative Y Coordinate. The X Coordinate will fall in between the 1st Vertex and the 3rd Vertex ———> call it (6 , -8)

3rd Vertex: (b , 0) will be the Vertex of the Triangle that has its X Coordinate closest to the Origin (0 , 0). Again, the Vertex/Point will lie on the X Axis as the 1st Point did. ——-> call if (4 , 0)

Take the Base as the straight Line across the X Axis from the 1st Vertex to the 3rd Vertex. The Length of the Base in Units is measured by the +Positive Distance between the 1st Vertex’s X Coordinate (a) and the 3rd Vertex’s X Coordinate (b)

Base = (a - b)

Height will be measured by the +Positive Perpendicular Distance from Vertex 2 to the Base on the X Axis. The Measure of the Height is given by the (+)Positive Value of the 2nd Vertex’s Y Coordinate (y)

Since y is a (-)Negative Value, we need to Negate the Value to make it Positive

Height = (-)y

Area of Triangle = (1/2) * (a - b) * (-)y

= (1/2) * (-ya + yb)

= (by - ay) / 2

-C-

Or, you can take the longer method and plug in the Values chosen above remembering that the Area of the Triangle must be (+)Positive

Posted from my mobile device

bumpbot · Answer

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In the coordinate plane, a triangle has vertices at (a,0),(b,0), and (

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