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Difficulty: 505-555 Level,    Coordinate Geometry,                      
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Re: In the coordinate plane, line k passes through the origin and has slop [#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: In the coordinate plane, line k passes through the origin and has slop [#permalink]
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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: In the coordinate plane, line k passes through the origin and has slop [#permalink]
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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

Show SpoilerMy 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: In the coordinate plane, line k passes through the origin and has slop [#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: In the coordinate plane, line k passes through the origin and has slop [#permalink]
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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In the coordinate plane, line k passes through the origin and has slop [#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.

Check Coordinate Geometry chapter of Math Book for more on this subject: https://gmatclub.com/forum/math-coordina ... 87652.html

Hope it helps.
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Re: In the coordinate plane, line k passes through the origin and has slop [#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: In the coordinate plane, line k passes through the origin and has slop [#permalink]
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Option C.
Origin=(0,0)
Let P1=(3,y)
Equation for line=>y-0=2(3-0)
y=6
Similarly,4-0=2(x-0)
x=2
x+y=8
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Re: In the coordinate plane, line k passes through the origin and has slop [#permalink]
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Slope of a line which passes through two point (x1,y1) and (x2,y2) is given by-
m=(y2-y1)/(x2-x1)

here m = 2
Here one point is Origin (0,0)
substituting the given values -
2 =(y-0)/(3-0)
y= 6

Similarly, for other point
2 =(4-0)/(x-0)
x =2


So x+y = 8

So correct answer is option C
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Re: In the coordinate plane, line k passes through the origin and has slop [#permalink]
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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

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

Problem Solving
Question: 102
Category: Algebra Simple coordinate geometry
Page: 74
Difficulty: 650


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Similar question : in-the-coordinate-plane-points-x-1-and-10-y-163337.html
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Re: In the coordinate plane, line k passes through the origin and has slop [#permalink]
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Hi All,

This question can be solved with a "brute force" approach, as long as you understand the Graphing vocabulary involved.

We're told that a line passes through the ORIGIN (meaning point 0,0) and has a SLOPE of 2 (meaning the Y-coordinate increases by 2 every time the X-coordiinate increases by 1).

Thus, we can list the first several points (starting at the Origin) without too much trouble:
(0, 0)
(1, 2)
(2, 4)
(3, 6)
(4, 8)
(5, 10)
Etc.

We're told that (3, Y) and (X, 4) are on this line. We're asked for the value of X+Y....

From the list (above), we can see that Y = 6 and that X = 2, so X+Y = 2+6 = 8

Final Answer:

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Re: In the coordinate plane, line k passes through the origin and has slop [#permalink]
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: In the coordinate plane, line k passes through the origin and has slop [#permalink]
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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: In the coordinate plane, line k passes through the origin and has slop [#permalink]
Bunuel wrote:
SOLUTION

In the coordinate plane, line k passes through the origin and has slope 2. If 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.


Hi Bunuel,

Why is the answer wrong if I do the question in the following way:

y=mx
y=2x
x+y = x+2x = 3x
Now from values given, 4 = 2x
x = 2
x+y = 3*2 = 6
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Re: In the coordinate plane, line k passes through the origin and has slop [#permalink]
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Hi nishatfarhat87,

In your work, you're confusing the VARIABLES X and Y with the X and Y co-ordinates on the line. Try doing your math again, but use these variables instead...

Y = 2X

(3, B) and (A, 4) are on the line. What is the value of A+B?

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Re: In the coordinate plane, line k passes through the origin and has slop [#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


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.

Answer: C
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Re: In the coordinate plane, line k passes through the origin and has slop [#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


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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