Can you show us your calculations?

That said, the distance formula isn't really necessary here, since the two given points have the same y-coordinate.

Cheers,
Brent

I used the formula to try to calculate the distance and it did not work. I was trying to use the formula to practice with.

I used the formula SQUARE ROOT OF [(x1-x2)² + (y1-y2)²]

(x1,y1) = (b,2)

(x1,y1) = (-2,2)

I got

3 = SQUARE ROOT OF (b+2)²
0 = b² + 4b + 4 -3
0 = b² + 4b + 1

And I got stuck here.

I couldn't solve for b because it wouldn't solve. I was trying to do something like 0 = (b + ?)(b + ?), but nothing would fit it to get the coordinates that are the answer.

I'm not sure what I'm doing wrong?

I did not understand how b=-5..Plz help

hi Reetika,
the points (b,2) and (-2,2) should tell you that these are coordinates as shown
Also since y- coordinates are 2 in both cases, the points are in a straight line parallel to x axis.

Now distance between the two points is 3 and this will totally depend on x coordinates..
so |b-(-2)|=3

two case..

|b+2|=3.....b+2=3.....b=1, so Column A>column B
|b+2|=3.....b+2=-3.....b=-5, so Column A<column B

D

thank you

If the distance from (b,2) to (-2,2) is 3, there are two possible cases:

CASE I:

For the above case, we get:
QUANTITY A: 1
QUANTITY B: 0
In this case, Quantity A is greater



CASE II:

For the above case, we get:
QUANTITY A: -5
QUANTITY B: 0
In this case, Quantity B is greater

Answer: D

Cheers,
Brent

Hi Brent GreenlightTestPrep, when using distance formula b = 1.
Not sure how is -5 come about and why do we need to consider negative option here since this is not involve any absolute or square root here?
Could you help clarify? Thanks Brent

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