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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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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

[Reveal] 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.
[Reveal] Spoiler: OA

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Re: X+Y= ? [#permalink]

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New post 10 Oct 2011, 08:55
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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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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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New post 10 Oct 2011, 10:07
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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

[Reveal] 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
Ques6.jpg [ 4.9 KiB | Viewed 18719 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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Re: OG PS: Line k with slope 2 [#permalink]

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New post 10 Oct 2011, 19:57
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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

[Reveal] 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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Re: OG PS: Line k with slope 2 [#permalink]

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New post 10 Oct 2011, 22:44
y=2x is the req. line
hence y=6 & x=2
so x+y=8
C
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Re: OG PS: Line k with slope 2 [#permalink]

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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)
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Re: OG PS: Line k with slope 2 [#permalink]

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New post 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?

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

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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.
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Re: OG PS: Line k with slope 2 [#permalink]

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

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New post 31 Mar 2012, 05:58
I made the same mistake as well.

(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.


Can someone explain why this is wrong. I couldn't understand.
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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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mymbadreamz wrote:
I made the same mistake as well.

(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.


Can someone explain why this is wrong. I couldn't understand.


This two posts address the same exact issue:
ln-the-coordinate-plane-line-k-passes-through-the-origin-121790.html#p1049905
ln-the-coordinate-plane-line-k-passes-through-the-origin-121790.html#p1050438
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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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

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

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New post 13 Sep 2015, 13:41
Bunuel wrote:
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.


I still don'Tget why we canot use the slope formula here ...`? y= 10-2x
The slope formula regards always 2 different points.. if we have 2 points we can build an equation of the line..

Using (0,0 and (3,y) we get the right answer, BUT using 3y and x,4 not
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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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New post 14 Sep 2015, 05:25
BrainLab wrote:
Bunuel wrote:
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.


I still don'Tget why we canot use the slope formula here ...`? y= 10-2x
The slope formula regards always 2 different points.. if we have 2 points we can build an equation of the line..

Using (0,0 and (3,y) we get the right answer, BUT using 3y and x,4 not


Have you checked this post: ln-the-coordinate-plane-line-k-passes-through-the-origin-121790.html#p1050438
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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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New post 14 Sep 2015, 06:00
Why is this wrong?[/quote]

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.[/quote]

I still don'Tget why we canot use the slope formula here ...`? y= 10-2x
The slope formula regards always 2 different points.. if we have 2 points we can build an equation of the line..

Using (0,0 and (3,y) we get the right answer, BUT using 3y and x,4 not[/quote]

Have you checked this post: ln-the-coordinate-plane-line-k-passes-through-the-origin-121790.html#p1050438[/quote]

Hi Bunuel, yes, I've already seen the reply from Karishma, but I still don't get it. What is the difference between using (3,y) & (x,4) and (3,y) & (0,0). Both points lay on the line. For exm. if we have coordinates (2,5) and (12,18) we can easily build an eyuation for this line, first finding the slope and then the y-intercept....
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Re: ln the coordinate plane, line k passes through the origin [#permalink]

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New post 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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New post 16 May 2016, 00:10
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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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Re: ln the coordinate plane, line k passes through the origin   [#permalink] 16 May 2016, 00:10

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