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In the normal x-y coordinate plane there are 4 points A(-1,3), B(3,3),

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In the normal x-y coordinate plane there are 4 points A(-1,3), B(3,3), [#permalink]

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[GMAT math practice question]

In the normal x-y coordinate plane there are 4 points A(-1,3), B(3,3), C(-1,7), and D(3,7). If a line passing through the origin bisects the area of the rectangle ABCD, what is the slope of the line?

A.5
B. 6
C. 7
D. 8
E. 9
[Reveal] Spoiler: OA

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Re: In the normal x-y coordinate plane there are 4 points A(-1,3), B(3,3), [#permalink]

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New post 13 Nov 2017, 00:48
Option A

The rectangle is a square of side 4. Area =16 units.
By drawing it on a paper the area can be divided into half by a line that makes a trapezium. Area of trapezium =1/2*h*(a+b) which in this case should be equal to 8.
Solving h(a+b)=16.
Here h=4. a+b=4. The only possible case is a=1.6 and b=2.4. When a=1.6 slope is 3/0.6. Hence 5


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The line should pass through the center of the rectangle ABCD.
Then the center is \((\frac{-1+3}{2}, \frac{3+7}{2})\) or \((1,5)\)
The slope of the line passing though \((0,0)\) and \((1,5)\) is \(\frac{5-0}{1-0}=5\).

Therefore, the answer is A.
Answer : A
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In the normal x-y coordinate plane there are 4 points A(-1,3), B(3,3),   [#permalink] 15 Nov 2017, 00:51
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