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# The figure shows that OABC is a parallelogram. What is the equation o

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Math Revolution GMAT Instructor
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The figure shows that OABC is a parallelogram. What is the equation o  [#permalink]

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13 Jan 2020, 00:16
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[GMAT math practice question]

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1.13PS.png [ 12.51 KiB | Viewed 493 times ]

The figure shows that $$OABC$$ is a parallelogram. What is the equation of the line passing through $$A$$ and $$C$$?

A. $$y = \frac{1}{10}x + \frac{3}{10}$$

B. $$y = \frac{5}{11}x + \frac{20}{11}$$

C. $$y = \frac{5}{12}x + \frac{3}{12 }$$

D. $$y = \frac{1}{13}x + \frac{3}{13}$$

E. $$y = \frac{5}{14}x + \frac{3}{14}$$

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" VP Joined: 20 Jul 2017 Posts: 1322 Location: India Concentration: Entrepreneurship, Marketing WE: Education (Education) Re: The figure shows that OABC is a parallelogram. What is the equation o [#permalink] ### Show Tags 13 Jan 2020, 05:08 1 MathRevolution wrote: [GMAT math practice question] Attachment: 1.13PS.png The figure shows that $$OABC$$ is a parallelogram. What is the equation of the line passing through $$A$$ and $$C$$? A. $$y = \frac{1}{10}x + \frac{3}{10}$$ B. $$y = \frac{5}{11}x + \frac{20}{11}$$ C. $$y = \frac{5}{12}x + \frac{3}{12 }$$ D. $$y = \frac{1}{13}x + \frac{3}{13}$$ E. $$y = \frac{5}{14}x + \frac{3}{14}$$ Let C = (a, 0) Given A = (7, 5), B = (3, 5) & O = (0, 0) Note that, Length of AB & OC are same --> $$\sqrt{(3 - 7)^2 + (5 - 5)^2} = \sqrt{(a - 0)^2 + (0 - 0)^2}$$ --> $$\sqrt{16 + 0} = \sqrt{a^2}$$ --> $$a = -4$$ [As "a" lies on the negative x-axis!] --> Point C = (-4, 0) Alternate method to find C: Note that point B is 4 units to the left of point A. So, Point C will also be 4 units to the left of O. --> C = (-4, 0) Slope of the line passing through A & C = $$\frac{0 - 5}{-4 - 7} = \frac{5}{11}$$ Only option B has slope = $$\frac{5}{11}$$ Option B Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8574 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: The figure shows that OABC is a parallelogram. What is the equation o [#permalink] ### Show Tags 15 Jan 2020, 00:25 => Since $$AB = CO = 4,$$ we have point $$C(-4, 0).$$ Then the slope of the line $$AC$$ is $$\frac{(5 - 0) }{ (7 - (-4))} = \frac{5}{11}.$$ The equation of the line $$AC$$ is $$y - 0 = (\frac{5}{11})(x - (-4))$$ or $$y = (\frac{5}{11})x + \frac{20}{11}.$$ Therefore, B is the answer. Answer: B _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: The figure shows that OABC is a parallelogram. What is the equation o  [#permalink]

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17 Jan 2020, 17:31
Easiest and quickest way would be to check the options satisfy point A(7,5) since the line passes through AC. and then see if multiple options satisfy it then go further to solve.
only B satisfies, hence B is the answer.
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Joined: 13 Jan 2020
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The figure shows that OABC is a parallelogram. What is the equation o  [#permalink]

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23 Jan 2020, 21:31
1
Let C = (a, 0)
Given A = (7, 5), B = (3, 5) & O = (0, 0)

Lengths of AB & OC are the same
$$\sqrt{(3−7)^2+(5−5)^2} = \sqrt{(a−0)^2+(0−0)^2}$$
a = −4
[As "a" lies on the negative x-axis!]
Point C = (-4, 0)

The slope of the line passing through A & C =$$\frac{ 0−5}{−4−7}=\frac{5}{11}$$
Only option B has slope = $$\frac{5}{11}$$

Option B
The figure shows that OABC is a parallelogram. What is the equation o   [#permalink] 23 Jan 2020, 21:31
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# The figure shows that OABC is a parallelogram. What is the equation o

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