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13 Jan 2020, 01:16
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
88% (02:02) correct
12%
(01:55)
wrong
based on 89
sessions
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[GMAT math practice question]
1.13PS.png [ 12.51 KiB | Viewed 2272 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. \(y = \frac{5}{14}x + \frac{3}{14}\)
Attachment:
1.13PS.png [ 12.51 KiB | Viewed 2272 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. \(y = \frac{5}{14}x + \frac{3}{14}\)
13 Jan 2020, 06:08
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MathRevolution
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
15 Jan 2020, 01:25
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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
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
