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

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09 Dec 2016, 05:44

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
_________________

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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GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions

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.

The equation of the line k, in slope-intercept form is y=2x so value of y must be double the value of x therefore for (3,y) we have y = 2*3 = 6 and for (x,4) we have x = 4/2 =2 => x + y = 6 + 2 = 8

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
_________________

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.

Slope intercept equation of the line is y=mx+c where c is the y intercept , as the line passes through origin thus we have c=0 Therefore our equation becomes y=mx Now two points are (3,y) and (x,4) m=2 then for (3,y) we have y=2*3=6 for (x,4) we have 4=2*x x=2 thus x+y=8

The answer is C
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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
_________________

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

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22 Aug 2017, 15:29

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?

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

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22 Aug 2017, 16:37

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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