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Math Revolution GMAT Instructor V
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A quadrilateral P has (1, 1), (3, 1), (3, 5) and (1, 5) as 4 vertices  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 52% (02:37) correct 48% (02:39) wrong based on 23 sessions

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[GMAT math practice question]

A quadrilateral $$P$$ has $$(1, 1), (3, 1), (3, 5)$$ and $$(1, 5)$$ as $$4$$ vertices and another quadrilateral $$Q$$ has $$(-1, -1), (-5, -1), (-5, -5)$$ and $$(-1, -5)$$ as $$4$$ vertices. A line divides these two quadrilaterals evenly at the same time. What is this line?

A. $$y = \frac{- 1}{6}x + \frac{5}{6 }$$

B. $$y = \frac{- 5}{6}x + \frac{1}{6}$$

C. $$y = \frac{ 6}{5}x + \frac{3}{5}$$

D. $$y = \frac{1}{3}x + \frac{5}{6}$$

E. $$y = \frac{- 1}{6}x + \frac{7}{5}$$

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A quadrilateral P has (1, 1), (3, 1), (3, 5) and (1, 5) as 4 vertices  [#permalink]

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By visualizing the two quadrilaterals, we find that one is in the first quadrant, and the other is in the fourth,
so for a line that can pass through them, the slope must be positive ---> this eliminates A,B,E

By trying D, the starting point of the line will be (3 , 1.833) which can't be possibly dividing the quadrilateral in the first quadrant.
By trying C, the starting point of the line will be (3 , 4.166) which can possibly do the trick
C
Attachments aaa.png [ 14.89 KiB | Viewed 330 times ]

Director  G
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Re: A quadrilateral P has (1, 1), (3, 1), (3, 5) and (1, 5) as 4 vertices  [#permalink]

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MathRevolution wrote:
[GMAT math practice question]

A quadrilateral $$P$$ has $$(1, 1), (3, 1), (3, 5)$$ and $$(1, 5)$$ as $$4$$ vertices and another quadrilateral $$Q$$ has $$(-1, -1), (-5, -1), (-5, -5)$$ and $$(-1, -5)$$ as $$4$$ vertices. A line divides these two quadrilaterals evenly at the same time. What is this line?

A. $$y = \frac{- 1}{6}x + \frac{5}{6 }$$

B. $$y = \frac{- 5}{6}x + \frac{1}{6}$$

C. $$y = \frac{ 6}{5}x + \frac{3}{5}$$

D. $$y = \frac{1}{3}x + \frac{5}{6}$$

E. $$y = \frac{- 1}{6}x + \frac{7}{5}$$

To divide a rectangle in half, a line must pass through the CENTER of the rectangle.

Center of P = (midpoint of the x-values, midpoint of the y-values) $$= (\frac{1+3}{2}, \frac{1+5}{2}) = (2, 3)$$
Center of Q = (midpoint of the x-values, midpoint of the y-values) $$= (\frac{-1+(-5)}{2}, \frac{-1+(-5)}{2}) = (-3, -3)$$

The correct answer must pass through the two centers (2, 3) and (-3, -3).
Slope of the line that passes through (2, 3) and (-3, -3) $$= \frac{∆y}{∆x} = \frac{-3-3}{-3-2} = \frac{6}{5}$$

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Re: A quadrilateral P has (1, 1), (3, 1), (3, 5) and (1, 5) as 4 vertices  [#permalink]

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=>

Attachment: 1.9ps(a).png [ 29.51 KiB | Viewed 206 times ]

The quadrilaterals are rectangles, and bisecting lines of rectangles pass through the center of the rectangles.
Thus we have to find the line passing through the centers of those two rectangles.
The centers of the rectangles are $$(2, 3)$$ and $$(-3, -3).$$

The slope of the line passing through $$(2, 3)$$ and $$(-3, -3)$$ is $$\frac{(3- (-3)) }{ (2 - (-3))}$$ $$=\frac{ 6}{5}$$.

The line passing through them is $$y – 3 = (\frac{6}{5})(x - 2)$$ or $$y = (\frac{6}{5})x + (\frac{3}{5}).$$

_________________ Re: A quadrilateral P has (1, 1), (3, 1), (3, 5) and (1, 5) as 4 vertices   [#permalink] 12 Jan 2020, 17:49
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# A quadrilateral P has (1, 1), (3, 1), (3, 5) and (1, 5) as 4 vertices  