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Re: The points (a+1, 1), (2a+1, 3) and (2a + 2, 2a) lie on the same straig [#permalink]
Hey, how did you factored \(2a^2-3a - 2=0\)?

Set equation equal to zero: \(2a^2-3a - 2=0\)
Factor: \((a-2)(2a+1)=0\)


GreenlightTestPrep wrote:
Carcass 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 = 2 or −1/2
(D) a = 2 or 1/2
(E) None of these.

Key property: If the three points are on the same straight line, then the slope between any two points will always be the SAME

In other words: The slope between (a+1, 1) and (2a+1, 3) = the slope between (2a+1, 3) and (2a + 2, 2a)

So, we can apply the slope formula to write: \(\frac{3 - 1}{(2a + 1) - (a + 1)} = \frac{2a - 3}{(2a+2) - (2a + 1)}\)

Simplify to get: \(\frac{2}{a} = \frac{2a - 3}{1}\)

Cross multiply to get: \((a)(2a-3) = (2)(1)\)

Simplify: \(2a^2-3a = 2\)

Set equation equal to zero: \(2a^2-3a - 2=0\)

Factor: \((a-2)(2a+1)=0\)

So, \(a = 2 \) or \(a = -\frac{1}{2}\)

Answer: C
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