In the x-y plane, the coordinates of three consecutive

BTGmoderatorLU wrote:In the x-y plane, the coordinates of three consecutive vertices of a rectangle are (2a, b), (-2a, b), (-2a, -3b). If a and b are integers with a<0 y b >0. Which of the following could be the coordinates of the fourth vertex?

A. (-3, -2)
B. (-2, 3)
C. (6, -9)
D. (-6, -9)
E. (-6, 9)

The OA is D.

Both B and D are possible.

For B, a = -1, b = 1
(-2,1), (2,1), (2,-3), and (-2,-3) for the coordinate (2a, -3b)(a square is a rectangle)

For D, a = -3, b = 3

Are the answer choices correct? Or am I missing something really obvious? Can anyone assist me please? Thanks in advance!

The axes of a two-dimensional Cartesian system divide the XY plane into four infinite regions, called quadrants, each bounded by two half-axes .

These are often numbered from I, II, III, and IV (numbered anti-clockwise):

• I (where the signs of the (x, y) coordinates are (+, +)),
  II (−,+),
  III (−,−), and
  IV (+,−).

We are given that the three vertices are: (2a, b), (-2a, b), and (-2a, -3b) such that a < 0 and b > 0.

Thus, we can rewrite the coordinates of the three vertices as (–2|a|, |b|), (2|a|, |b|), and (2|a|, –3|b|).

Since

• of (–2|a|, |b|), the x coordinate is negative and the y coordinate is positive, the vertice must lie in the II quadrant;
  of (2|a|, |b|), the x, as well as y coordinate, is positive, the vertice must lie in the I quadrant;
  of (2|a|, –3|b|), the x coordinate is positive and the y coordinate is negative, the vertice must lie in the IV quadrant

So, three of the four vertices lie in three different quadrants (I, II and IV), thus, the fourth vertice would lie in the III quadrant.

We know that a point that lies in the III quadrant has its x, as well as y coordinate negative, thus, out of the five options, A or D can be the correct answer.

The coordinates of the fourth vertice would be (–2|a|, –3|b|). We are given that a and b are integers, thus, Option A cannot be correct answer as –2|a| ≠ –3, else a would not be an integer.

Thus, the correct answer is D.

Let's check if a and b are integers.

We have –2|a| = –6 => |a| = 3, an integer; similarly, –3|b| = –9 => |b| = 3, an integer.

The correct answer: D

Hope this helps!

-Jay
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BTGmoderatorLU wrote:In the x-y plane, the coordinates of three consecutive vertices of a rectangle are (2a, b), (-2a, b), (-2a, -3b). If a and b are integers with a<0 y b >0. Which of the following could be the coordinates of the fourth vertex?

A. (-3, -2)
B. (-2, 3)
C. (6, -9)
D. (-6, -9)
E. (-6, 9)

We see two of the vertices have x-coordinate of -2a and one has 2a; so the fourth vertex must have an x-coordinate of 2a. Similarly, we see two of the vertices have y-coordinate of b and one has -3b; so the fourth vertex must have a y-coordinate of -3b. Thus, the coordinates of the fourth vertex must be (2a, -3b).

Since a and b are integers with a<0 and b >0, we see that 2a must be a negative multiple of 2 and -3b must be a negative multiple of 3. The only coordinates that fit this condition are (-6, -9).

Answer: D .

Hi BTGmoderatorLU,

We're told that in the x-y plane, the coordinates of three consecutive vertices of a rectangle are (2A, B), (-2A, B), (-2A, -3B), A and B are both INTEGERS with A < 0 and B > 0. We're asked which of the following COULD be the coordinates of the fourth vertex? TESTing VALUES is a great approach for this prompt, but you made a small error in one of your calculations.

In your example, A = -1, B = 1, the three given co-ordinates would be (-2, 1), (2, 1), and (2, -3), so the missing coordinate would need to be (-2, -3). That is NOT what's stated in Answer B. Your work for Answer D is correct.

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Rich