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In the xy-coordinate system, if (a,b) and (a + 3, b + k) are two point

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In the xy-coordinate system, if (a,b) and (a + 3, b + k) are two point [#permalink]

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New post 17 Dec 2017, 23:14
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In the xy-coordinate system, if (a,b) and (a + 3, b + k) are two points on the line defined by the equation x = 3y – 1, then k =

(A) 9
(B) 3
(C) 7/3
(D) 1
(E) 1/3
[Reveal] Spoiler: OA

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Re: In the xy-coordinate system, if (a,b) and (a + 3, b + k) are two point [#permalink]

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New post 18 Dec 2017, 05:41
Bunuel wrote:
In the xy-coordinate system, if (a,b) and (a + 3, b + k) are two points on the line defined by the equation x = 3y – 1, then k =

(A) 9
(B) 3
(C) 7/3
(D) 1
(E) 1/3


x = 3y – 1

i.e. y = (1/3)x + (1/3)

General Equation of Line is y = mx+c

where m = slope

Comparing the two equations here we get,

SLope of line, m = (1/3) = [(b+k)-b] / [(a+3)-a]

i.e. (1/3) = k/3

i.e. k = 1

Answer: option D
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Re: In the xy-coordinate system, if (a,b) and (a + 3, b + k) are two point [#permalink]

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New post 23 Apr 2018, 05:50
Bunuel wrote:
In the xy-coordinate system, if (a,b) and (a + 3, b + k) are two points on the line defined by the equation x = 3y – 1, then k =

(A) 9
(B) 3
(C) 7/3
(D) 1
(E) 1/3


A is a trap answer

First convert the equation in \(y = mx + c\) form.

\(x = 3y - 1\)

\(y = \frac{x}{3} +\frac{1}{3}\) ------ (I)

Slope = \(\frac{1}{3}\)

\(b + k = \frac{1}{3}(a+3) + \frac{1}{3}\)

\(b = \frac{1}{3}(a+3) + \frac{1}{3} - k\) ------ (II)

\(b =\frac{1}{3}(a) + \frac{1}{3}\) ----- (III)

Equating II and III we get

\(k = 1\)

(D)
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Re: In the xy-coordinate system, if (a,b) and (a + 3, b + k) are two point   [#permalink] 23 Apr 2018, 05:50
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