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In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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 21 Apr 2008, 16:49
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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-7, 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 [#permalink]
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21 Apr 2008, 17:11
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x=3*y-7
convert to (this step is not necessary but I do it out of habit)
y=(x-7)/3
substitute a and b for x and y
1. b=(a-7)/3
b+k=((a+3)-7)/3
substitute equation 1 for b
(a-7)/3+k=((a+3)-7)/3
(a-7)/3+k=(a-4)/3
a-7+3k=a-4
3k=3
k=1
D
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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21 Apr 2008, 20:29
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alimad 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 - 7, then k =
9
3
7/3
1
1/3
Please provide explaination. Thanks given : y = 1/3x + 7/3 Slope => k/3 = 1/3 i.e K =1
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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21 Apr 2008, 20:45
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alimad 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 - 7, then k =
9
3
7/3
1
1/3
Please provide explaination. Thanks x = 3y - 7 y = 1/3 (x) + 7/3 [(b+k) - b]/ [(a+3) - a] = 1/3 k/3 = 1/3 k = 1 D.
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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21 Apr 2008, 21:30
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alimad 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 - 7, then k =
9
3
7/3
1
1/3
Please provide explaination. Thanks slope = (y1-y2)/(x1-x2) = (b+k-b) / (a+3-a) = k/3 y = x/3 + 7/3, so slope = 1/3 k/3 = 1/3 k = 1
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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27 Jun 2012, 15:36
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Substitute a & b in place of x & y resp... in the eqn. a-3b+7=0------(i) Substitute a+3 & b+k in the place of x & y resp...we'll get a+3=3(b+k)-7 or, a-3b-3k+10=0-------(ii) points on the same line will satisfy the equation so , equating (i)&(ii) a-3b-3k+10=a-3b+7 k=1 Ans- D
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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27 Jun 2012, 23:10
thevenus wrote: Substitute a & b in place of x & y resp... in the eqn.
a-3b+7=0------(i)
Substitute a+3 & b+k in the place of x & y resp...we'll get
a+3=3(b+k)-7 or,
a-3b-3k+10=0-------(ii)
Or note here itself that a - 3b + 7 = 0 so 3 - 3k = 0 giving you k = 1
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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27 Jun 2012, 23:12
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thevenus 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-7, then k=?
A. 9
B. 3
C. 7/3
D. 1
E. 1/3 My answer is (D).
Here's my approach:
We are given two points: (a,b) and (a+3,b+k)
We are given an equation of the line: x = 3y - 7
Next step is to convert the equation of the line into slope-intercept form. We will have Y = x/3 + 7/3. Don't forget this step because you might fall into the trap and decide that the slope of the line is 3. This makes the question very GMAT-esque. 
So the slope of the line is 1/3
Now we know that the equation of the slope of the line is given by:
slope = (y2 - y1)/(x2 - x1) ---> Remember, the slope is just "rise" over "run."
That's why we have:
1/3 = [ (b+k) - b ] / [ (a+3) - a]
The two b's will cancel each other in the numerator and so will the two a's in the denominator
We will get 1/3 = k / 3
so 3 / 3 = k
k = 1
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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27 Jun 2012, 23:39
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Hi,
Slope of the line x=3y-7 is 1/3
so, we can equate the slope of the line to the slope of the points = \(\frac {y_2-y_1}{x_2-x_1}\)
or \(\frac {(b+k)-(b)}{(a+3)-a} = \frac 13\)
or \(\frac k3 = \frac 13\)
or k=1,
Answer (D),
Regards,
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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16 Feb 2015, 07:57
Maple wrote: x=3*y-7
convert to (this step is not necessary but I do it out of habit)
y=(x-7)/3
substitute a and b for x and y
1. b=(a-7)/3
b+k=((a+3)-7)/3
substitute equation 1 for b
(a-7)/3+k=((a+3)-7)/3
(a-7)/3+k=(a-4)/3
a-7+3k=a-4
3k=3
k=1
D x=3*y-7 is not y=(x-7)/3instead it is y=(x+7)/3 or y=(1/3)x+(7/3) since we know (a,b) and (a+3,b+k) belong to the same line, they must have the same slope, 1/3. [(b+k)-b]/[(a+3)-a]=1/3 k/3=1/3 k=1
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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16 Feb 2015, 19:16
Hi All, This question can be solved by TESTing VALUES: We're given the equation of a line (X = 3Y - 7) and we're told that two points (A, B) and (A+3, B+K) are on this line. We're asked for the value of K. In graphing questions, it sometimes helps to "visualize" the line better if you write the equation in "slope-intercept" format: X = 3Y - 7 3Y = X + 7 Y = X/3 + 7/3 For the first co-ordinate, let's try to keep things simple... X = 0 Y = 7/3 So... A = 0 B = 7/3 For the second co-ordinate, we have ADD 3 to X.... X = 3 Y = 10/3 So.... A+3 = 3 B+K = 10/3 We know from the first co-ordinate that B = 7/3, so K = 3/3 = 1 Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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14 Jul 2016, 05:25
alimad 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-7, then k=?
A. 9
B. 3
C. 7/3
D. 1
E. 1/3 Our equation is x=3y-7 Lets quickly make it a point intercept form ==> 3y-7=x==>3y=x+7==>\(y=\frac{x}{3}+\frac{7}{3}\) Now we can see the coefficient of x is \(\frac{1}{3}\), which by definition is the slope and we know slope =\(\frac{y2-y1}{x2-x1}\)==> \(\frac{1}{3}=\frac{b+k-b}{a+3-a}\)==> 1/3=k/3 therefore k=1 Answer is B
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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30 Oct 2017, 13:56
alimad 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-7, then k=?
A. 9
B. 3
C. 7/3
D. 1
E. 1/3 Recall that an ordered pair represents a pair of x and y coordinates. Substituting the values from the first ordered pair (a,b) into the equation, we can create the following equation: a = 3b - 7 Substituting the values from the second ordered pair for x and y into the same equation, we have: a + 3 = 3(b + k) - 7 → a + 3 = 3b + 3k - 7 If we subtract the first equation from the second, we have: 3 = 3k 1 = k Answer: D
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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26 Jan 2018, 14:27
EMPOWERgmatRichC wrote: Hi All, This question can be solved by TESTing VALUES: We're given the equation of a line (X = 3Y - 7) and we're told that two points (A, B) and (A+3, B+K) are on this line. We're asked for the value of K. In graphing questions, it sometimes helps to "visualize" the line better if you write the equation in "slope-intercept" format: X = 3Y - 7 3Y = X + 7 Y = X/3 + 7/3 For the first co-ordinate, let's try to keep things simple... X = 0 Y = 7/3 So... A = 0 B = 7/3 For the second co-ordinate, we have ADD 3 to X.... X = 3 Y = 10/3 So.... A+3 = 3 B+K = 10/3 We know from the first co-ordinate that B = 7/3, so K = 3/3 = 1 Final Answer: GMAT assassins aren't born, they're made, Rich Hi Rich, I know the "7/3" came from rewriting to slope-intercept form, but how did you get from "B = 7/3" to "B+K = 10/3"? I know we added 3 to X, but are we adding 3 to Y too?
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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26 Jan 2018, 15:03
Hi OCDianaOC, Both co-ordinates have to 'fit' the equation Y = X/3 + 7/3 The first co-ordinate is (A, B).... and I TESTed VALUES and used (0, 7/3) to define that co-ordinate. Remember: A = 0 and B = 7/3 The second co-ordinate is (A+3, B+K).... notice how that's the SAME A and B from the first co-ordinate. Thus, we have to add 3 to A (so 3+0 = 3) and plug in X=3 into the equation to get the value of the Y.... When X=3.... Y = X/3 + 7/3 Y = (3/3) + 7/3) Y = 10/3 Since the second co-ordinate is (A+3, B+K), our prior work makes the co-ordinate (3, 10/3). From the prior work, we see that B = 7/3... B + K = 10/3 (7/3) + K = 10/3 K = 10/3 - 7/3 = 3/3 = 1 GMAT assassins aren't born, they're made, Rich
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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21 Feb 2018, 13:40
gmatsaga wrote: thevenus 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-7, then k=?
A. 9
B. 3
C. 7/3
D. 1
E. 1/3 My answer is (D).
Here's my approach:
We are given two points: (a,b) and (a+3,b+k)
We are given an equation of the line: x = 3y - 7
Next step is to convert the equation of the line into slope-intercept form. We will have Y = x/3 + 7/3. Don't forget this step because you might fall into the trap and decide that the slope of the line is 3. This makes the question very GMAT-esque. 
So the slope of the line is 1/3
Now we know that the equation of the slope of the line is given by:
slope = (y2 - y1)/(x2 - x1) ---> Remember, the slope is just "rise" over "run."
That's why we have:
1/3 = [ (b+k) - b ] / [ (a+3) - a]
The two b's will cancel each other in the numerator and so will the two a's in the denominator
We will get 1/3 = k / 3
so 3 / 3 = k
k = 1
How did you figure out that slope is 1/3 from here Y = x/3 + 7/3 slope is x/3 but how should i know value of X ? 
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In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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21 Feb 2018, 21:08
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alimad 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-7, then k=?
A. 9
B. 3
C. 7/3
D. 1
E. 1/3 \(x = 3y - 7\) Rewrite in slope-intercept form \(y = mx + b\)m = slope, b = y-intercept, thus: \(3y = x + 7\) => \(y = \frac{1}{3}x + \frac{7}{3}\) Slope = \(\frac{1}{3}\) Slope is also \(\frac{rise}{run}=\frac{(y_2 - y_1)}{(x_2 - x_1)}\) We have x-and y-coordinates for two points: (a,b) and (a+3,b+k) Set the slope equation equal to the slope value \(\frac{(b+k)-b}{(a+3)-a}=\frac{1}{3}\) \(\frac{b+k-b}{a+3-a}=\frac{1}{3}\) \(\frac{k}{3}=\frac{1}{3}\) \(3k = 3\) \(k = 1\) Answer D
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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21 Feb 2018, 21:13
dave13 wrote: gmatsaga wrote: thevenus 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-7, then k=?
A. 9
B. 3
C. 7/3
D. 1
E. 1/3 My answer is (D).
Here's my approach:
We are given two points: (a,b) and (a+3,b+k)
We are given an equation of the line: x = 3y - 7
Next step is to convert the equation of the line into slope-intercept form. We will have Y = x/3 + 7/3. Don't forget this step because you might fall into the trap and decide that the slope of the line is 3. This makes the question very GMAT-esque.
So the slope of the line is 1/3
Now we know that the equation of the slope of the line is given by:
slope = (y2 - y1)/(x2 - x1) ---> Remember, the slope is just "rise" over "run."
That's why we have:
1/3 = [ (b+k) - b ] / [ (a+3) - a]
The two b's will cancel each other in the numerator and so will the two a's in the denominator
We will get 1/3 = k / 3
so 3 / 3 = k
k = 1
How did you figure out that slope is 1/3 from here Y = x/3 + 7/3 slope is x/3 but how should i know value of X ? The equation of a line is y = mx + c where m is the slope and c is the y-intercept. So the equation looks like this y = 2x + 4 (m = 2, c = 4) y = x/3 + 5 (m = 1/3, c = 5) etc For more, check out this post: http://www.veritasprep.com/blog/2010/12 ... he-graphs/
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Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two
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