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

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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
3
1
1
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
2
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

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
1
2
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

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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Posts: 1324
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
1
1
1
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

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
2
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
1
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

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
3
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,

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)/3
instead 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
3
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

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ Director Joined: 04 Jun 2016 Posts: 563 GMAT 1: 750 Q49 V43 Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink] ### Show Tags 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 _________________ Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired. Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2823 Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink] ### Show Tags 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 _________________ # Jeffrey Miller Head of GMAT Instruction Jeff@TargetTestPrep.com 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Intern Joined: 16 Oct 2017 Posts: 37 Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink] ### Show Tags 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? EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 14209 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two [#permalink] ### Show Tags 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 _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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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

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
1
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

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

$$\frac{b+k-b}{a+3-a}=\frac{1}{3}$$

$$\frac{k}{3}=\frac{1}{3}$$

$$3k = 3$$

$$k = 1$$

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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

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  [#permalink]

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03 Sep 2018, 06:58
Top Contributor
1
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

Key Concept: If a point lies ON a line, then the coordinates (x and y) of that point must SATISFY the equation of that line.

Given equation: x = 3y - 7
One point ON the line is (a, b)
So, we can write: a = 3b - 7

Another point ON the line is (a + 3, b + k)
So, we can write: a + 3 = 3(b + k) - 7
Expand: a + 3 = 3b + 3k - 7
Subtract 3 from both sides to get: a = 3b + 3k - 10

We now two equations:
a = 3b + 3k - 10
a = 3b - 7

Subtract the bottom equation from the top equation to get: 0 = 3k - 3
Add 3 to both sides: 3 = 3k
Solve: k = 1

Cheers,
Brent
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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 Nov 2018, 07:35
another solution since x=3y-7 is increasing line ie if x increases then y increasing(x will increase by y's constant y will increase by x's constant)
hence D
Re: In the xy-coordinate system,if (a,b) and (a+3, b+k) are two   [#permalink] 14 Nov 2018, 07:35

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