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The points (a+1, 1), (2a+1, 3) and (2a + 2, 2a) lie on the same straig

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Senior Manager
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Joined: 25 Dec 2018
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GMAT Date: 02-18-2019
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The points (a+1, 1), (2a+1, 3) and (2a + 2, 2a) lie on the same straig  [#permalink]

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New post 05 Mar 2019, 23:19
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A
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Question Stats:

41% (02:10) correct 59% (01:39) wrong based on 17 sessions

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The points (a+1, 1), (2a+1, 3) and (2a + 2, 2a) lie on the same straight line if

A) a= -1 or 2
B) a = 2 or 1
C) a=2or−12
D) a=2or12
E) None of these.
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Re: The points (a+1, 1), (2a+1, 3) and (2a + 2, 2a) lie on the same straig  [#permalink]

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New post 05 Mar 2019, 23:48
1
mangamma wrote:
The points (a+1, 1), (2a+1, 3) and (2a + 2, 2a) lie on the same straight line if

A) a= -1 or 2
B) a = 2 or 1
C) a=2or−12
D) a=2or12
E) None of these.


Three points are collinear if the value of area of triangle formed by the three points is zero.

So, \((a+1)(3-2a)–1(2a+1–2a–2) +2a(2a+1)–3(2a+2)=0\)
Or, \(2a^2-3a-2=0\)
Or, \(2a^2-4a+a-2=0\)
Or, 2a(a-2)+1(a-2)=0
Or, (a-2)(2a+1)=0
Or a=2, -1/2

Ans. (C)
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Re: The points (a+1, 1), (2a+1, 3) and (2a + 2, 2a) lie on the same straig  [#permalink]

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New post 12 Mar 2019, 04:37
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1
mangamma wrote:
The points (a+1, 1), (2a+1, 3) and (2a + 2, 2a) lie on the same straight line if

A) a= -1 or 2
B) a = 2 or 1
C) a=2or−12
D) a=2or12
E) None of these.


If the points lie in a straight line then their slope will be equal.
Taking slopes of point 2 and 3 and equating with the slope of point 1 and 2
\(\frac{2a-3}{1} =\frac{2}{a}\)
solving this gives a= 2 or a= \(-\frac{1}{2}\)

P.S. option C should be corrected it's not -12 but -\(\frac{1}{2}\)
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Re: The points (a+1, 1), (2a+1, 3) and (2a + 2, 2a) lie on the same straig   [#permalink] 12 Mar 2019, 04:37
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The points (a+1, 1), (2a+1, 3) and (2a + 2, 2a) lie on the same straig

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