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In the xycoordinate 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 xycoordinate system, if (a,b) and (a+3, b+k) are two points on the line defined by the equation x=3y7, then k=? A. 9 B. 3 C. 7/3 D. 1 E. 1/3
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Re: In the xycoordinate 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*y7 convert to (this step is not necessary but I do it out of habit) y=(x7)/3
substitute a and b for x and y 1. b=(a7)/3
b+k=((a+3)7)/3 substitute equation 1 for b (a7)/3+k=((a+3)7)/3 (a7)/3+k=(a4)/3 a7+3k=a4 3k=3 k=1
D



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Re: In the xycoordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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alimad wrote: In the xycoordinate 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 xycoordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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alimad wrote: In the xycoordinate 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 xycoordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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alimad wrote: In the xycoordinate 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 = (y1y2)/(x1x2) = (b+kb) / (a+3a) = k/3 y = x/3 + 7/3, so slope = 1/3 k/3 = 1/3 k = 1



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Re: In the xycoordinate 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. a3b+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, a3b3k+10=0(ii) points on the same line will satisfy the equation so , equating (i)&(ii) a3b3k+10=a3b+7 k=1 Ans D
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Re: In the xycoordinate 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. a3b+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, a3b3k+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 xycoordinate 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 xycoordinate system, if (a,b) and (a+3, b+k) are two points on the line defined by the equation x=3y7, 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 slopeintercept 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 GMATesque.
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 xycoordinate 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=3y7 is 1/3 so, we can equate the slope of the line to the slope of the points = \(\frac {y_2y_1}{x_2x_1}\) or \(\frac {(b+k)(b)}{(a+3)a} = \frac 13\) or \(\frac k3 = \frac 13\) or k=1,
Answer (D),
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Re: In the xycoordinate 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*y7 convert to (this step is not necessary but I do it out of habit) y=(x7)/3
substitute a and b for x and y 1. b=(a7)/3
b+k=((a+3)7)/3 substitute equation 1 for b (a7)/3+k=((a+3)7)/3 (a7)/3+k=(a4)/3 a7+3k=a4 3k=3 k=1
D x=3*y7 is not y=(x7)/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 xycoordinate 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 "slopeintercept" format: X = 3Y  7 3Y = X + 7 Y = X/3 + 7/3 For the first coordinate, let's try to keep things simple... X = 0 Y = 7/3 So... A = 0 B = 7/3 For the second coordinate, we have ADD 3 to X.... X = 3 Y = 10/3 So.... A+3 = 3 B+K = 10/3 We know from the first coordinate 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 xycoordinate 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 xycoordinate system, if (a,b) and (a+3, b+k) are two points on the line defined by the equation x=3y7, then k=?
A. 9 B. 3 C. 7/3 D. 1 E. 1/3 Our equation is x=3y7 Lets quickly make it a point intercept form ==> 3y7=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 =\(\frac{y2y1}{x2x1}\)==> \(\frac{1}{3}=\frac{b+kb}{a+3a}\)==> 1/3=k/3 therefore k=1 Answer is B
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Re: In the xycoordinate 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 xycoordinate system, if (a,b) and (a+3, b+k) are two points on the line defined by the equation x=3y7, 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 xycoordinate 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 "slopeintercept" format: X = 3Y  7 3Y = X + 7 Y = X/3 + 7/3 For the first coordinate, let's try to keep things simple... X = 0 Y = 7/3 So... A = 0 B = 7/3 For the second coordinate, we have ADD 3 to X.... X = 3 Y = 10/3 So.... A+3 = 3 B+K = 10/3 We know from the first coordinate 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 slopeintercept 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 xycoordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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26 Jan 2018, 15:03
Hi OCDianaOC, Both coordinates have to 'fit' the equation Y = X/3 + 7/3 The first coordinate is (A, B).... and I TESTed VALUES and used (0, 7/3) to define that coordinate. Remember: A = 0 and B = 7/3 The second coordinate is (A+3, B+K).... notice how that's the SAME A and B from the first coordinate. 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 coordinate is (A+3, B+K), our prior work makes the coordinate (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 xycoordinate 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 xycoordinate system, if (a,b) and (a+3, b+k) are two points on the line defined by the equation x=3y7, 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 slopeintercept 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 GMATesque.
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 xycoordinate system,if (a,b) and (a+3, b+k) are two [#permalink]
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alimad wrote: In the xycoordinate system, if (a,b) and (a+3, b+k) are two points on the line defined by the equation x=3y7, then k=?
A. 9 B. 3 C. 7/3 D. 1 E. 1/3 \(x = 3y  7\) Rewrite in slopeintercept form \(y = mx + b\)m = slope, b = yintercept, 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 xand ycoordinates for two points: (a,b) and (a+3,b+k) Set the slope equation equal to the slope value \(\frac{(b+k)b}{(a+3)a}=\frac{1}{3}\) \(\frac{b+kb}{a+3a}=\frac{1}{3}\) \(\frac{k}{3}=\frac{1}{3}\) \(3k = 3\) \(k = 1\) Answer D
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Re: In the xycoordinate 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 xycoordinate system, if (a,b) and (a+3, b+k) are two points on the line defined by the equation x=3y7, 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 slopeintercept 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 GMATesque.
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 yintercept. 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 ... hegraphs/
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