Author 
Message 
Manager
Joined: 18 Aug 2010
Posts: 90

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



Math Expert
Joined: 02 Sep 2009
Posts: 39576

Re: m18#24 [#permalink]
Show Tags
31 Jan 2011, 11:50
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: mathcoordinategeometry87652.html (chapter "Lines in Coordinate Geometry").
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7438
Location: Pune, India

Re: m18#24 [#permalink]
Show Tags
14 Feb 2013, 21:21
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 coordinate 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 = 1x, 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 unintuitive. 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.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Manager
Joined: 20 Dec 2010
Posts: 168
Location: Stockholm, Sweden

Re: m18#24 [#permalink]
Show Tags
30 Jan 2011, 16:19
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) (01/2)/(10) = 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 yintercept. The second coordinate (0,1/2) tells us that when x = 0 y = 1/2, which is the yintercept or 'm' in the formula y = kx+m where m is the yintercept 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 = 1x
_________________
12/2010 GMATPrep 1 620 (Q34/V41) 01/2011 GMATPrep 2 640 (Q42/V36) 01/2011 GMATPrep 3 700 (Q47/V39) 02/2011 GMATPrep 4 710 (Q48/V39) 02/2011 MGMAT CAT 1 650 (Q46/V32) 02/2011 MGMAT CAT 2 680 (Q46/V36) 02/2011 MGMAT CAT 3 710 (Q45/V41)
Last edited by Mackieman on 31 Jan 2011, 04:34, edited 3 times in total.



Manager
Joined: 18 Aug 2010
Posts: 90

Re: m18#24 [#permalink]
Show Tags
31 Jan 2011, 01:56
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



Manager
Joined: 18 Aug 2010
Posts: 90

Re: m18#24 [#permalink]
Show Tags
31 Jan 2011, 01:58
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 \(2y = 1  x\) .



Manager
Joined: 18 Aug 2010
Posts: 90

Re: m18#24 [#permalink]
Show Tags
31 Jan 2011, 02:16
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)
(01/2)/(10) = 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 yintercept.
The first coordinate (1,0) tells us that when y = 0 x = 1, which is the yintercept 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?



Manager
Joined: 20 Dec 2010
Posts: 168
Location: Stockholm, Sweden

Re: m18#24 [#permalink]
Show Tags
31 Jan 2011, 04:33
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)
(01/2)/(10) = 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 yintercept.
The first coordinate (1,0) tells us that when y = 0 x = 1, which is the yintercept 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)
_________________
12/2010 GMATPrep 1 620 (Q34/V41) 01/2011 GMATPrep 2 640 (Q42/V36) 01/2011 GMATPrep 3 700 (Q47/V39) 02/2011 GMATPrep 4 710 (Q48/V39) 02/2011 MGMAT CAT 1 650 (Q46/V32) 02/2011 MGMAT CAT 2 680 (Q46/V36) 02/2011 MGMAT CAT 3 710 (Q45/V41)



Manager
Joined: 18 Aug 2010
Posts: 90

Re: m18#24 [#permalink]
Show Tags
31 Jan 2011, 11:57
I saw the link. VERY IMPRESSIVE!!!! + kudo from me
GREAT JOB !!!



Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 520
Location: India
GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)

Re: m18#24 [#permalink]
Show Tags
13 Feb 2013, 01:31
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: mathcoordinategeometry87652.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. .
_________________
hope is a good thing, maybe the best of things. And no good thing ever dies.
Who says you need a 700 ?Check this out : http://gmatclub.com/forum/whosaysyouneeda149706.html#p1201595
My GMAT Journey : http://gmatclub.com/forum/endofmygmatjourney149328.html#p1197992



Intern
Joined: 02 Aug 2013
Posts: 16

For every point (a,b) lying on line 1, point (b,a) lies on line [#permalink]
Show Tags
28 Dec 2013, 23:57
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, 23:57






