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

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Joined: 02 Sep 2009
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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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17 Dec 2017, 23:14
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Difficulty:

35% (medium)

Question Stats:

67% (00:47) correct 33% (01:08) wrong based on 128 sessions

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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

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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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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

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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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23 Apr 2018, 05:50
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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

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 &nbs [#permalink] 23 Apr 2018, 05:50
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