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# ln the coordinate plane, line k passes through the origin

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ln the coordinate plane, line k passes through the origin [#permalink]

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10 Oct 2011, 08:30
1
17
00:00

Difficulty:

25% (medium)

Question Stats:

74% (01:28) correct 26% (01:27) wrong based on 812 sessions

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ln the coordinate plane, line k passes through the origin and has slope 2. lf points (3,y) and (x,4) are on line k, then x+y =

(A) 3.5
(B) 7
(C) 8
(D) 10
(E) 14

Spoiler: :: My Take
i used the method y2-y1/x1-x2 =2

4-y/x-3 =2

As line passes through origin y=2x
Upon solving both the equation i get x+y=7.5

OOps some different method required.

Guys is this the trick in Coordinate geometry that we generally have to put points in line and chk.instead of making equations with the points given.

Pls comment

@Fluke ---i searched this question but didn't found.

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10 Oct 2011, 08:55
5
I think we have to go with general equation of line with two points.

y-y1/y2-y1 = x-x1/x2-x1

And substituting the values:

y-4 = 2(x-x1) => x1 =2 (as (0,0) is on the line).
4-y1=2(x-3) => y1= 6 (as (0,0) is on the line).

x+y = 8; Ans:C.
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Re: OG PS: Line k with slope 2 [#permalink]

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10 Oct 2011, 09:05
11
3
GMATD11 wrote:
ln the coordinate plane, line k passes through the origin and has slope 2. lf points (3,y) and (x,4) are on line k, then x+y =

(A) 3.5
(B) 7
(C) 8
(D) 10
(E) 14

Sol:
y=mx+c
m=Slope=2
c=intersection on y-axis=0
y=2x

The line passes through (3,y)
y=2*3=6

The line also passes through (x,4)
4=2*x;
x=2;

x+y=2+6=8

Ans: "C"
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Re: OG PS: Line k with slope 2 [#permalink]

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10 Oct 2011, 10:07
22
14
GMATD11 wrote:
ln the coordinate plane, line k passes through the origin and has slope 2. lf points (3,y) and (x,4) are on line k, then x+y =

(A) 3.5
(B) 7
(C) 8
(D) 10
(E) 14

Spoiler: :: My Take
i used the method y2-y1/x1-x2 =2

4-y/x-3 =2

As line passes through origin y=2x
Upon solving both the equation i get x+y=7.5

OOps some different method required.

Guys is this the trick in Coordinate geometry that we generally have to put points in line and chk.instead of making equations with the points given.

Pls comment

@Fluke ---i searched this question but didn't found.

To find the line k, knowing a point through which it passes and the line's slope is sufficient.
Since it passes through origin and has slope 2, this is what it will look like:
Attachment:

Ques6.jpg [ 4.9 KiB | Viewed 24511 times ]

What is the meaning of slope? It means for every increase of 1 unit in x coordinate, y coordinate increases by 2 units. When x coordinate increases by 3 units (from the point (0,0)), y coordinate increases by 6 units. Hence the line passes through (3, 6). y must be 6.
When y coordinate increases by 4 units (from (0,0)), x coordinate must have increased by 2 units. The line must pass through (2, 4). x must be 2.
x+y = 2+6 = 8
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Joined: 01 Feb 2011 Posts: 686 Re: OG PS: Line k with slope 2 [#permalink] ### Show Tags 10 Oct 2011, 19:57 3 x and y in those points (x,4) and y in (3,y) can be confusing . you ran into this situation because you are trying to solve two equations y = 2x and 2x+y=10. Thats not correct x and y in 2x+y = 10 refer to x in (x,4) and y in (3,y) and belong to two different points where as y=2x is the generic equation . here x and y belong to same point. to avoid confusion i would say you can replace the points as (3,p) and (q,4) and rephrase the question as p+q=? equation of the line passing through origin and slope 2 => y = 2x then you can compare slopes slope of (3,p) and (0,0) = 2 => p/3 = 2 => p=6 slope of (q,4) and (0,0) = 2 => 4/q = 2 => q = 2 => p+q = 8 GMATD11 wrote: ln the coordinate plane, line k passes through the origin and has slope 2. lf points (3,y) and (x,4) are on line k, then x+y = (A) 3.5 (B) 7 (C) 8 (D) 10 (E) 14 Spoiler: :: My Take i used the method y2-y1/x1-x2 =2 4-y/x-3 =2 As line passes through origin y=2x Upon solving both the equation i get x+y=7.5 OOps some different method required. Guys is this the trick in Coordinate geometry that we generally have to put points in line and chk.instead of making equations with the points given. Pls comment @Fluke ---i searched this question but didn't found. Manager Joined: 16 Dec 2009 Posts: 69 GMAT 1: 680 Q49 V33 WE: Information Technology (Commercial Banking) Re: OG PS: Line k with slope 2 [#permalink] ### Show Tags 24 Nov 2011, 22:58 3 Points (3,y) and origin lies on the same line and have a slope of 2. So (y-0)/(3-0)=2 or y=6 Points (x,4) and origin lies on the same line and have a slope of 2. So (4-0)/(x-0)=2 or x=2 So x+y=8 .. Answer choice (C) _________________ If Electricity comes from Electrons , Does Morality come from Morons ?? If you find my post useful ... then please give me kudos ...... h(n) defined as product of even integers from 2 to n Number N divided by D leaves remainder R Ultimate list of MBA scholarships for international applicants Intern Joined: 16 Jan 2012 Posts: 4 Re: OG PS: Line k with slope 2 [#permalink] ### Show Tags 25 Feb 2012, 15:24 I got this wrong and I cant figure what is wrong with my approach (y2-y1)/(x2-x1)=m (slope) so we have 2 points x,4 and 3,y so we get the eq (y-4)/(3-x)=2. simplifying this we get : y= 10-2x so I get the value of y=4 and x=3 and hence the sum 7. Why is this wrong? Math Expert Joined: 02 Sep 2009 Posts: 46287 Re: OG PS: Line k with slope 2 [#permalink] ### Show Tags 25 Feb 2012, 15:50 5 1 Ashamock wrote: I got this wrong and I cant figure what is wrong with my approach (y2-y1)/(x2-x1)=m (slope) so we have 2 points x,4 and 3,y so we get the eq (y-4)/(3-x)=2. simplifying this we get : y= 10-2x so I get the value of y=4 and x=3 and hence the sum 7. Why is this wrong? Welcome to GMAT Club. Below is an answer to your question: There are infinitely many values of x and y possible to satisfy y=10-2x. For any value of x there exist some y for which y=10-2x holds true (and vise versa). For example: x=1, y=8 or x=2, y=6, or x=0.5, y=9, ... Also notice that you won't be able to find the value of x+y without one more piece of information given in the stem, namely information saying that line k passes through the origin. There are several approaches discussed above how to handle this problem, I'd offer one more. ln the coordinate plane, line k passes through the origin and has slope 2. lf points (3,y) and (x,4) are on line k, then x+y = (A) 3.5 (B) 7 (C) 8 (D) 10 (E) 14 Any line which passes through the origin has a form of $$y=mx$$, since $$m=2$$ (the slope of a line) then we have that the equation of our line is $$y=2x$$. Now, if we substitute the coordinates of two points we'll get: For point (3,y) --> $$y=2*3=6$$; For point (x,4) --> $$4=2x$$ --> $$x=2$$; $$x+y=8$$. Answer: C. Check Coordinate Geometry chapter of Math Book for more on this subject: math-coordinate-geometry-87652.html Hope it helps. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8102 Location: Pune, India Re: OG PS: Line k with slope 2 [#permalink] ### Show Tags 27 Feb 2012, 04:53 2 Ashamock wrote: I got this wrong and I cant figure what is wrong with my approach (y2-y1)/(x2-x1)=m (slope) so we have 2 points x,4 and 3,y so we get the eq (y-4)/(3-x)=2. simplifying this we get : y= 10-2x so I get the value of y=4 and x=3 and hence the sum 7. Why is this wrong? Let me point out one thing I noticed: How did you get y = 4 and x = 3? The equation you get is y= 10-2x. Mind you, here y = y2 and x = x1 i.e. they are y and x co-ordinates of different points. y = 10 - 2x can give you the value of y + 2x i.e. equal to 10 but how do you get the value of (y+x)? You can go from here and find the answer if you consider the info that the line passes through the center. Say, the 2 points on the line that you are considering are (0, 0) and (x, 4) (y2-y1)/(x2-x1)=m (4 - 0)/(x - 0) = 2 x = 2 Now you can plug x = 2 in y = 10 - 2x to get y = 6 Their sum 2+6 = 8 Hope you understand that these x and y stand for particular values. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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08 Jul 2013, 01:10
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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04 Sep 2013, 02:37
1
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

Y = mx + c where m=slope and c = Y intercept

We have that m = slope = 2 AND c = Y Intercept = 0 (We know like passes thru origin

Hence The equation of line is y = (2)x + 0 ------> y = 2x

point (3,y) ------> y = 6
point (x,4) -------> 4 = 2x ------> x = 2

Hence x + y 8
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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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15 May 2016, 22:01
I'm stumped!

what is wrong in the approach below?

y= 2x+c (we know that c=0). Thus,
y=2x........equation 1

next,
(4-y)/(x-3)= 2
Thus, 10= 2x+y....equation 2

From equation 1 & equation 2, I found x= 5/2 & y= 5.
Thus x+y= 7.5

I'm not understanding what is wrong in this process!
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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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16 May 2016, 00:10
1
kanav06 wrote:
I'm stumped!

what is wrong in the approach below?

y= 2x+c (we know that c=0). Thus,
y=2x........equation 1

next,
(4-y)/(x-3)= 2
Thus, 10= 2x+y....equation 2

From equation 1 & equation 2, I found x= 5/2 & y= 5.
Thus x+y= 7.5

I'm not understanding what is wrong in this process!

y = 2x is the equation of the line - fine.

But these two - (3,y) and (x,4) are points. Here y and x are specific co-ordinate points.

Think of them instead as points (3, a) and (b, 4). You need to find a + b.
You will get 10 = 2a + b by equating it to slope but how will you solve for a + b?

The equation of the line is y = 2x
So a = 2*3 = 6
Also, 4 = 2*b
b = 2

So a + b = 6 + 2 = 8
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ln the coordinate plane, line k passes through the origin [#permalink]

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09 Dec 2016, 06:44
VeritasPrepKarishma wrote:
kanav06 wrote:
I'm stumped!

what is wrong in the approach below?

y= 2x+c (we know that c=0). Thus,
y=2x........equation 1

next,
(4-y)/(x-3)= 2
Thus, 10= 2x+y....equation 2

From equation 1 & equation 2, I found x= 5/2 & y= 5.
Thus x+y= 7.5

I'm not understanding what is wrong in this process!

y = 2x is the equation of the line - fine.

But these two - (3,y) and (x,4) are points. Here y and x are specific co-ordinate points.

Think of them instead as points (3, a) and (b, 4). You need to find a + b.
You will get 10 = 2a + b by equating it to slope but how will you solve for a + b?

The equation of the line is y = 2x
So a = 2*3 = 6
Also, 4 = 2*b
b = 2

So a + b = 6 + 2 = 8

But these two - (3,y) and (x,4) are points. Here y and x are specific co-ordinate points.
So, why slope formula for two points can not be used? Both points lie on the same line. Slope of any two points of a line is same. Need details.

Initially, I also tried to solve like kanav06
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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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12 Dec 2016, 18:19
1
GMATD11 wrote:
ln the coordinate plane, line k passes through the origin and has slope 2. lf points (3,y) and (x,4) are on line k, then x+y =

(A) 3.5
(B) 7
(C) 8
(D) 10
(E) 14

We are given that line k passes through the origin (or the point (0,0)) and has a slope of 2. Since slope = (change in y)/(change in x), we can create the following equation using the coordinates (0,0) and (3,y):

2 = (y - 0)/(3 - 0)

2 = y/3

y = 6

We can use the slope equation again, this time using the coordinates (0,0) and (x,4):

2 = (4 - 0)/(x - 0)

2 = 4/x

2x = 4

x = 2

Thus, x + y = 6 + 2 = 8.

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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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16 Aug 2017, 08:56
1
hi Bunuel VeritasPrepKarishma

Is this approach correct:
A slope of a line is given by y = mx+ c
Since the line passes through origin (given) if we substitute (0,0) in above equation, we get c=0
Now we also know from question stem that points (3,y) and (x,4) are on line k, hence below two equations can be inferred:
y = 2 * 3 + 0 so we get y = 6
4 = 2 * x + 0 so we get x = 2
So x + y = 8

Yes this approach is perfectly correct... and this is the only shortest way..

You can go like this..
Slope is 2. So , m =2 .Line passes through origin. So, c= 0 .. Line is Y=2x
y = 2*3 = 6; 4=2*x ; x+y = 6+2 = 8
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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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22 Aug 2017, 16:29
Quote:
y = 2x is the equation of the line - fine.

But these two - (3,y) and (x,4) are points. Here y and x are specific co-ordinate points.

Think of them instead as points (3, a) and (b, 4). You need to find a + b.
You will get 10 = 2a + b by equating it to slope but how will you solve for a + b?

The equation of the line is y = 2x
So a = 2*3 = 6
Also, 4 = 2*b
b = 2

So a + b = 6 + 2 = 8

I got the question at hand right (C), but I am uncomfortable as to how I got it right -- it felt like guessing.

here is what I did, step by step:

1. y = mx + b
2. y = 2x + 0 given slope = 2, and the origin
3. y = 2x
After step three, I dead ended and did not know how to use "y = 2x" to solve for the missing points.

So, I just looked at the problem from a rise over run POV

2 = rise / run

2 = y-4 / 3-x <--- right here, I essentially guessed by plugging in numbers until the equation worked.
2 = (6) - 4 / 3 - (2) <--- I plugged in 6 for Y and 2 for X because I knew it would give me 2; however, I know that this way is essientially guessing.

What is a simplified solution to the problem, following the steps I followed?
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ln the coordinate plane, line k passes through the origin [#permalink]

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22 Aug 2017, 17:37
1
CyberStein wrote:
Quote:
y = 2x is the equation of the line - fine.

But these two - (3,y) and (x,4) are points. Here y and x are specific co-ordinate points.

Think of them instead as points (3, a) and (b, 4). You need to find a + b.
You will get 10 = 2a + b by equating it to slope but how will you solve for a + b?

The equation of the line is y = 2x
So a = 2*3 = 6
Also, 4 = 2*b
b = 2

So a + b = 6 + 2 = 8

I got the question at hand right (C), but I am uncomfortable as to how I got it right -- it felt like guessing.

here is what I did, step by step:

1. y = mx + b
2. y = 2x + 0 given slope = 2, and the origin
3. y = 2x
After step three, I dead ended and did not know how to use "y = 2x" to solve for the missing points.

So, I just looked at the problem from a rise over run POV

2 = rise / run

2 = y-4 / 3-x <--- right here, I essentially guessed by plugging in numbers until the equation worked.
2 = (6) - 4 / 3 - (2) <--- I plugged in 6 for Y and 2 for X because I knew it would give me 2; however, I know that this way is essientially guessing.

What is a simplified solution to the problem, following the steps I followed?

CyberStein
You had it! I think you just forgot that you already found the relationship between y and x (in the equation) -- and, once you have the linear function in the form of the equation, you can find one coordinate if you have the other.

The missing x and y have nothing to do with one another except that they are on the same line. The question is asking for one more trivial step after you do the work: can you track correctly and add?

All you need to do: use the line equation you found, take one pair of coordinates at a time, and plug in the given value to find the missing one.

For (3,y). We know THIS x is 3. What is its matching y-coordinate? Plug in the given x-value to find the y-value for this pair. We need the "missing" y in (3,y).

y = 2x
Given: x = 3. Plug it in.
y = (2)(3)
y = 6
y equals 6 when x equals 3. The pair is now (3, 6).
If you were graphing, and went to x equals 3, you would go up to y equals 6.

Now the other pair. We are given (x,4). We need x.

y = 2x
Given: y = 4. Plug it in.
4 = 2x
4/2 = x
x = 2
For this pair we have (2,4).

This pair is a little odd because we are used to thinking of y as a function of x, and not the other way around. But we already have the relationship between y and x. It's in the equation you wrote.

The question at the end is trivial, meaning, it has nothing to do with the relationship between the x and the y -- except what a randomly picked x-value and a randomly picked y value, from that line, add up to.

The question is, my rewrite: "[Now that you have figured out the line equation, and figured out particular values for missing coordinates from two different pairs], what does [that] x + [that] y equal?"

We got x equals 2 on the one hand, and y equals 6, on the other hand. Add them.

2 + 6 = 8

Maybe you were thinking too hard, that's all. I should all have such problems!

I couldn't quite tell where you were going wrong, so I guessed. Does that help?
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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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06 Feb 2018, 15:56
Top Contributor
GMATD11 wrote:
ln the coordinate plane, line k passes through the origin and has slope 2. lf points (3,y) and (x,4) are on line k, then x+y =

(A) 3.5
(B) 7
(C) 8
(D) 10
(E) 14

It often helps to write equations for lines in the form y = mx + b (this is called slope y-intercept form), where m = the slope of the line, and b = the y-intercept of the line.

The question tells us that the line has slope 2. So, m = 2
The question also tells us that the line passes through the origin (0,0). So, the y-intercept is 0, which means b = 0
So, the equation of the line is y = 2x + 0, or just y = 2x

If the point (3,y) is on the line, then its coordinates must satisfy the equation y = 2x
So, plug x=3 and y=y into the equation to get y = (2)(3) = 6

If the point (x,4) is on the line, then its coordinates must satisfy the equation y = 2x
So, plug x=x and y=4 into the equation to get 4 = 2x, which means x = 2

So, x + y = 2 + 6
= 8
= C

Cheers,
Brent
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Re: ln the coordinate plane, line k passes through the origin   [#permalink] 06 Feb 2018, 15:56
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