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

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

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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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New post 13 Jan 2020, 05:08
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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
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New post 15 Jan 2020, 00:25
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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
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Re: The figure shows that OABC is a parallelogram. What is the equation o  [#permalink]

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