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

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

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)

Area of the triangle with vertices (x1, y1) , (x2, y2) , (x3, y3) =\(\frac{1}{2}\) [ (x2y3 -x3y2 ) - (x1y3 - x3y1 ) + (x1y2 -x2y1) ]

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

So area of triangle is = \(\frac{1}{2}\) [ (by -0 ) - (ay - 0 ) + ( 0 - 0) ]

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

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.

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.

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 \(\frac{(by−ay)}{2} = -y*\frac{(a-b)}{2} = -negative*positive = positive*positive = positive\).

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

how did you get the base? a-b

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

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