Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 29 May 2015, 12:34

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m18#24

Author Message
Manager
Joined: 17 Aug 2010
Posts: 90
Followers: 1

Kudos [?]: 5 [0], given: 22

m18#24 [#permalink]  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 ?

(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 $$2y = 1 - x$$ .

i cant understand what the formula used here for building equation line 2. Can somebody explain? what is the Formula ?
thank you
 Kaplan Promo Code Knewton GMAT Discount Codes Veritas Prep GMAT Discount Codes
Math Expert
Joined: 02 Sep 2009
Posts: 27535
Followers: 4328

Kudos [?]: 42607 [2] , given: 6050

Re: m18#24 [#permalink]  31 Jan 2011, 10:50
2
KUDOS
Expert's post
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 $$2y = 1 - x$$ .

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").
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5573
Location: Pune, India
Followers: 1375

Kudos [?]: 7013 [1] , given: 181

Re: m18#24 [#permalink]  14 Feb 2013, 20:21
1
KUDOS
Expert's post
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..

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.
_________________

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
Followers: 2

Kudos [?]: 33 [0], given: 5

Re: m18#24 [#permalink]  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
_________________

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, 03:34, edited 3 times in total.
Manager
Joined: 17 Aug 2010
Posts: 90
Followers: 1

Kudos [?]: 5 [0], given: 22

Re: m18#24 [#permalink]  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
Manager
Joined: 17 Aug 2010
Posts: 90
Followers: 1

Kudos [?]: 5 [0], given: 22

Re: m18#24 [#permalink]  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 $$2y = 1 - x$$ .
Manager
Joined: 17 Aug 2010
Posts: 90
Followers: 1

Kudos [?]: 5 [0], given: 22

Re: m18#24 [#permalink]  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?
Manager
Joined: 20 Dec 2010
Posts: 168
Location: Stockholm, Sweden
Followers: 2

Kudos [?]: 33 [0], given: 5

Re: m18#24 [#permalink]  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)
_________________

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: 17 Aug 2010
Posts: 90
Followers: 1

Kudos [?]: 5 [0], given: 22

Re: m18#24 [#permalink]  31 Jan 2011, 10:57
I saw the link. VERY IMPRESSIVE!!!! + kudo from me

GREAT JOB !!!
Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 548
Location: India
GMAT 1: 640 Q43 V34
GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)
Followers: 2

Kudos [?]: 37 [0], given: 562

Re: m18#24 [#permalink]  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 $$2y = 1 - x$$ .

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..
_________________

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/who-says-you-need-a-149706.html#p1201595

My GMAT Journey : end-of-my-gmat-journey-149328.html#p1197992

Intern
Joined: 02 Aug 2013
Posts: 17
Followers: 1

Kudos [?]: 4 [0], given: 1

For every point (a,b) lying on line 1, point (b,-a) lies on line [#permalink]  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
Display posts from previous: Sort by

# m18#24

Moderators: Bunuel, WoundedTiger

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.