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m18#24

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m18#24 [#permalink] New post 30 Jan 2011, 12:55
For every point (a, b) lying on line 1, point (b, -a) lies on line 2. If the equation of line 1 is y = 2x + 1 , what is the equation of line 2 ?

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* y = \frac{1}{2} + \frac{x}{2}
* 2y = 1 - x
* \frac{x + y}{2} = -1
* y = \frac{x}{2} - 1
* x = 2y + 1

Find two points on line 2 and use their coordinates to build the line's equation. Points (0, 1) and (-\frac{1}{2}, 0) on line 1 correspond to points (1, 0) and (0, \frac{1}{2}) on line 2. The equation of line 2 is \frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1} or 2y = 1 - x .
The correct answer is B.


i cant understand what the formula used here for building equation line 2. Can somebody explain? what is the Formula ?
thank you
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Re: m18#24 [#permalink] New post 30 Jan 2011, 15:19
The formula for finding the slope of a line is (change in y)/(change in x)

In your case, two coordinates (1,0) and (0,1/2) where (x,y)

(0-1/2)/(1-0) = -1/2 which is the slope of the line. The common formula for a line is y = kx + m where k is the slope and m is the y-intercept.

The second coordinate (0,1/2) tells us that when x = 0 y = 1/2, which is the y-intercept or 'm' in the formula y = kx+m where m is the y-intercept and 'k' is the slope. Plugging in the slope (-1/2) and the intercept (1/2) into the formula gives us y = 1/2-(x/2)

multiplying by 2 on both sides:
2y = 1-x
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Last edited by Mackieman on 31 Jan 2011, 03:34, edited 3 times in total.
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Re: m18#24 [#permalink] New post 31 Jan 2011, 00:56
thanks . yet im actually wondering what is the formula used here that comes to direct equation. ???

The equation of line 2 is \frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1} or 2y = 1 - x
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Re: m18#24 [#permalink] New post 31 Jan 2011, 00:58
sorry, here is the equation of my interest :

The equation of line 2 is \frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1} or 2y = 1 - x .
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Re: m18#24 [#permalink] New post 31 Jan 2011, 01:16
Mackieman wrote:
The formula for finding the slope of a line is (change in y)/(change in x)

In your case, two coordinates (1,0) and (0,1/2) where (x,y)

(0-1/2)/(1-0) = -1/2 which is the slope of the line. The common formula for a line is y = kx + m where k is the slope and m is the y-intercept.

The first coordinate (1,0) tells us that when y = 0 x = 1, which is the y-intercept or 'm' in the formula y = kx+m where m is the intercept and 'k' is the slope. Pluggin in the slope (-1/2) and the intercept (1) into the formula gives ut y = 1-(x/2)


y- intercept is at x=0 as i know. so y= 1/2 when x=0 Am i missing something?
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Re: m18#24 [#permalink] New post 31 Jan 2011, 03:33
tinki wrote:
Mackieman wrote:
The formula for finding the slope of a line is (change in y)/(change in x)

In your case, two coordinates (1,0) and (0,1/2) where (x,y)

(0-1/2)/(1-0) = -1/2 which is the slope of the line. The common formula for a line is y = kx + m where k is the slope and m is the y-intercept.

The first coordinate (1,0) tells us that when y = 0 x = 1, which is the y-intercept or 'm' in the formula y = kx+m where m is the intercept and 'k' is the slope. Pluggin in the slope (-1/2) and the intercept (1) into the formula gives ut y = 1-(x/2)


y- intercept is at x=0 as i know. so y= 1/2 when x=0 Am i missing something?


Sorry, you are correct, I had a rough day yesterday ;-) I hope it makes sense now.
(edited my post above)
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Re: m18#24 [#permalink] New post 31 Jan 2011, 10:50
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tinki wrote:
For every point (a, b) lying on line 1, point (b, -a) lies on line 2. If the equation of line 1 is y = 2x + 1 , what is the equation of line 2 ?

(C) 2008 GMAT Club - m18#24

* y = \frac{1}{2} + \frac{x}{2}
* 2y = 1 - x
* \frac{x + y}{2} = -1
* y = \frac{x}{2} - 1
* x = 2y + 1

Find two points on line 2 and use their coordinates to build the line's equation. Points (0, 1) and (-\frac{1}{2}, 0) on line 1 correspond to points (1, 0) and (0, \frac{1}{2}) on line 2. The equation of line 2 is \frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1} or 2y = 1 - x .
The correct answer is B.


i cant understand what the formula used here for building equation line 2. Can somebody explain? what is the Formula ?
thank you


Check this: math-coordinate-geometry-87652.html (chapter "Lines in Coordinate Geometry").
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Re: m18#24 [#permalink] New post 31 Jan 2011, 10:57
I saw the link. VERY IMPRESSIVE!!!! + kudo from me

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Re: m18#24 [#permalink] New post 13 Feb 2013, 00:31
Bunuel wrote:
tinki wrote:
For every point (a, b) lying on line 1, point (b, -a) lies on line 2. If the equation of line 1 is y = 2x + 1 , what is the equation of line 2 ?

(C) 2008 GMAT Club - m18#24

* y = \frac{1}{2} + \frac{x}{2}
* 2y = 1 - x
* \frac{x + y}{2} = -1
* y = \frac{x}{2} - 1
* x = 2y + 1

Find two points on line 2 and use their coordinates to build the line's equation. Points (0, 1) and (-\frac{1}{2}, 0) on line 1 correspond to points (1, 0) and (0, \frac{1}{2}) on line 2. The equation of line 2 is \frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1} or 2y = 1 - x .
The correct answer is B.


i cant understand what the formula used here for building equation line 2. Can somebody explain? what is the Formula ?
thank you


Check this: math-coordinate-geometry-87652.html (chapter "Lines in Coordinate Geometry").



bunuel/Karishma

This is how I solved and it worked but I don't know why it worked :D Please enligthen me about why it worked...

I plugged in (a,b) in line 1 equeation to get b=2a+1..

then I started plugging in (b,-a) in the answer choices.. the 2nd answer choice resulted in b=2a+1.. and hence I marked B..
I am not sure why this worked.. Please help. .
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Re: m18#24 [#permalink] New post 14 Feb 2013, 20:21
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Sachin9 wrote:
I plugged in (a,b) in line 1 equeation to get b=2a+1..

then I started plugging in (b,-a) in the answer choices.. the 2nd answer choice resulted in b=2a+1.. and hence I marked B..
I am not sure why this worked.. Please help. .


Responding to a pm:

a and b stand for two numbers which define a co-ordinate on the plane. When you say that (a, b) lies on y = 2x+1, it means the relation between a and b is b = 2a + 1. e.g. if a = 0, b = 1; if a = 1, b = 3... At the end of the day, a line is nothing but a depiction of how one variable changes with another. A line just shows you the relation between 2 variables.

If (b, -a) lies on a line 2y = 1-x, this is just a different way of expressing the same relation between the two numbers a and b.
a and b are the same set of numbers (i.e. if a = 0, b = 1; if a = 1, b = 3...)
So after manipulating the equation a little, you are bound to get b = 2a + 1 only.

As you figured out, the approach is a little un-intuitive. When I looked at the problem, I actually solved it exactly the same way except that I took numbers rather than a and b.

I said, if (a, b) lies on y = 2x + 1, if a = 1, b = 3.
So (3, -1) must lie on the new equation of the line. When I put (3, -1) in the options, I see that only (B) satisfies.
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For every point (a,b) lying on line 1, point (b,-a) lies on line [#permalink] New post 28 Dec 2013, 22:57
it took me a while to get this

so what i did was create a value for x then sub it into line 1 to find out what y is. now that you have (a,b) create the point (b,-a).

now what you do is sub is points of line 2 into various equations until you find one that makes sense.
For every point (a,b) lying on line 1, point (b,-a) lies on line   [#permalink] 28 Dec 2013, 22:57
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