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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 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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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.

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) --> \(4=2x\) --> \(x=2\);

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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19 Oct 2014, 07:11

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

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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) --> \(4=2x\) --> \(x=2\);

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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Share some Kudos, if my posts help you. Thank you !

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) --> \(4=2x\) --> \(x=2\);

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

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

Show Tags

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) --> \(4=2x\) --> \(x=2\);

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]

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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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