GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 12 Dec 2019, 01:56 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # M18-24

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 59685

### Show Tags

2
8 00:00

Difficulty:   45% (medium)

Question Stats: 72% (02:23) correct 28% (02:28) wrong based on 168 sessions

### HideShow timer Statistics

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 ?

A. $$y = \frac{1}{2} + \frac{x}{2}$$
B. $$2y = 1 - x$$
C. $$\frac{x + y}{2} = -1$$
D. $$y = \frac{x}{2} - 1$$
E. $$x = 2y + 1$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59685

### Show Tags

Official Solution:

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 ?

A. $$y = \frac{1}{2} + \frac{x}{2}$$
B. $$2y = 1 - x$$
C. $$\frac{x + y}{2} = -1$$
D. $$y = \frac{x}{2} - 1$$
E. $$x = 2y + 1$$

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 $$2y = 1 - x$$.

_________________
Intern  Joined: 05 Sep 2014
Posts: 9

### Show Tags

5
Hi all

Not sure if it is correct, but it worked, so I would just like to make sure that my thought process was correct. We have y=2x + 1. And (a,b) is on that line, therefore (a,b) = (x,y); the stem also states (b,-a) lies on line 2 for every (a,b), therefore line 2 is (-y,x). My reasoning is as follows: Line 1 = (x,y) = (a,b) [x=a; y=b]; Line 2 = (-b,a) = (-y,x) thus
x=2(-y) + 1 --> 2y = 1 - x

Thanks
Intern  Joined: 08 Feb 2015
Posts: 4

### Show Tags

2
I think this can also be solved by way of negative reciprocals since the two lines are perpendicular.

From the premise it seems that is the case since every point (a,b) is (b, -a) on the other line (pointing to a different direction and y and x being reversed.

Hence the answer will be the equation with slope = - 1/2. (To find the negative reciprocal of a number, flip the number over (invert) and negate that value)
Current Student S
Joined: 08 Feb 2017
Posts: 80
Concentration: Entrepreneurship, Marketing
GMAT 1: 660 Q48 V33 GMAT 2: 710 Q49 V38 GPA: 3.32

### Show Tags

Bunuel wrote:
Official Solution:

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 ?

A. $$y = \frac{1}{2} + \frac{x}{2}$$
B. $$2y = 1 - x$$
C. $$\frac{x + y}{2} = -1$$
D. $$y = \frac{x}{2} - 1$$
E. $$x = 2y + 1$$

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 $$2y = 1 - x$$.

Could you please elaborate on why $$\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}$$ or $$2y = 1 - x$$?
Math Expert V
Joined: 02 Sep 2009
Posts: 59685

### Show Tags

Evgart wrote:
Bunuel wrote:
Official Solution:

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 ?

A. $$y = \frac{1}{2} + \frac{x}{2}$$
B. $$2y = 1 - x$$
C. $$\frac{x + y}{2} = -1$$
D. $$y = \frac{x}{2} - 1$$
E. $$x = 2y + 1$$

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 $$2y = 1 - x$$.

Could you please elaborate on why $$\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}$$ or $$2y = 1 - x$$?

$$\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}$$;

$$\frac{y }{\frac{1}{2}} = \frac{x - 1}{- 1}$$;

$$2y=-x+1$$;

$$2y = 1-x$$.
_________________
Current Student S
Joined: 08 Feb 2017
Posts: 80
Concentration: Entrepreneurship, Marketing
GMAT 1: 660 Q48 V33 GMAT 2: 710 Q49 V38 GPA: 3.32

### Show Tags

Bunuel wrote:
Evgart wrote:
Bunuel wrote:
Official Solution:

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 ?

A. $$y = \frac{1}{2} + \frac{x}{2}$$
B. $$2y = 1 - x$$
C. $$\frac{x + y}{2} = -1$$
D. $$y = \frac{x}{2} - 1$$
E. $$x = 2y + 1$$

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 $$2y = 1 - x$$.

Could you please elaborate on why $$\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}$$ or $$2y = 1 - x$$?

$$\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}$$;

$$\frac{y }{\frac{1}{2}} = \frac{x - 1}{- 1}$$;

$$2y=-x+1$$;

$$2y = 1-x$$.

Thank you, I can understand that:-)
My question is more about why:

$$\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}$$.[/quote]
Math Expert V
Joined: 02 Sep 2009
Posts: 59685

### Show Tags

Evgart wrote:

Thank you, I can understand that:-)
My question is more about why:

$$\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}$$.
[/quote]

This is the way to find equation of a line given two points. Check coordinate geometry chapter of math book for more: https://gmatclub.com/forum/math-coordin ... 87652.html
_________________
Intern  B
Joined: 03 May 2014
Posts: 26

### Show Tags

Bunuel : where am I going wrong?
Line equation passing through two points (1,0) and (0,1/2) - line 2.
Equation of line => y-y1/x-x1 = y1-y2/x1-x2
y-0/x-1 = (0-1/2)/(1-0)
y/x-1 = -1/2
y = -1/2(x-1)
y= -1/2x + 1/2
Math Expert V
Joined: 02 Sep 2009
Posts: 59685

### Show Tags

ManSab wrote:
Bunuel : where am I going wrong?
Line equation passing through two points (1,0) and (0,1/2) - line 2.
Equation of line => y-y1/x-x1 = y1-y2/x1-x2
y-0/x-1 = (0-1/2)/(1-0)
y/x-1 = -1/2
y = -1/2(x-1)
y= -1/2x + 1

I was not able to decipher your way above because of poor formatting. Anyway below is correct way:

$$\frac{y - 0}{\frac{1}{2} - 0} = \frac{x - 1}{0 - 1}$$;

$$\frac{y }{\frac{1}{2}} = \frac{x - 1}{- 1}$$;

$$2y=-x+1$$;

$$2y = 1-x$$.
_________________
Intern  B
Joined: 03 May 2014
Posts: 26

### Show Tags

@Buneul - not able to figure out your steps, hence asking your review for my steps. This time with better format.
Equation of line formula = $$\frac{y−y1}{x-x1}$$ =$$\frac{y1-y2}{x1-x2}$$
Using your example line points (1,0) and (0,1/2)

$$\frac{y-0}{x-1}$$ =$$\frac{0 - \frac{1}{2}}{1-0}$$

$$\frac{y}{x-1}$$=$$\frac{-1}{2}$$

y =$$\frac{-1}{2}(x-1)$$
y=$$\frac{-1}{2}x+\frac{1}{2}$$

Math Expert V
Joined: 02 Sep 2009
Posts: 59685

### Show Tags

ManSab wrote:
@Buneul - not able to figure out your steps, hence asking your review for my steps. This time with better format.
Equation of line formula = $$\frac{y−y1}{x-x1}$$ =$$\frac{y1-y2}{x1-x2}$$
Using your example line points (1,0) and (0,1/2)

$$\frac{y-0}{x-1}$$ =$$\frac{0 - \frac{1}{2}}{1-0}$$

$$\frac{y}{x-1}$$=$$\frac{-1}{2}$$

y =$$\frac{-1}{2}(x-1)$$
y=$$\frac{-1}{2}x+\frac{1}{2}$$

No mistake there.

Multiply by 2: $$2y = -x + 1$$ or $$2y = 1 - x$$.
_________________
Intern  B
Joined: 03 May 2014
Posts: 26

### Show Tags

Thank you Bunuel.!
Intern  B
Joined: 25 Jan 2017
Posts: 4

### Show Tags

1
My approach:

Let a = 2. Therefore x = 2. Substitute in L1's eqn and solve for y. Thus, y = 5 (i.e. b = 5).
This means there is a point on L2 with the coordinates (5, -2). Substitute either of these points in the answer choices and identify the equation that, when solved, gives you the other point.
Manager  G
Status: Studying 4Gmat
Joined: 02 Jan 2016
Posts: 238
Location: India
Concentration: Strategy, Entrepreneurship
GMAT 1: 590 Q37 V33 GPA: 4
WE: Law (Manufacturing)

### Show Tags

Hi Bunuel,

Can you explain how did you get (0,1) on line 1 and (-1/2,0) on line 2.
Math Expert V
Joined: 02 Sep 2009
Posts: 59685

### Show Tags

hero_with_1000_faces wrote:
Hi Bunuel,

Can you explain how did you get (0,1) on line 1 and (-1/2,0) on line 2.

The equation of line 1 is y = 2x + 1.

If x = 0, then y = 2*0 + 1 = 1. So, point (0, 1) is on line y = 2x + 1. The stem says: for every point (a,b) lying on line 1, point (b,−a) lies on line 2. Point (a = 0, b = 1) is on line 1, so point (b = 1, -a = 0) is on line 2.

If y = 0, then 0 = 2x + 1 --> x = -1/2. So, point (-1/2, 0) is on line y = 2x + 1. The stem says: for every point (a,b) lying on line 1, point (b,−a) lies on line 2. Point (a= -1/2, b = 0) is on line 1, then point (b = 0, -a = 1/2) is on line 2.
_________________
Manager  B
Joined: 20 Jun 2018
Posts: 72
Location: India

### Show Tags

Hey Bunuel is there a directory kinda thing which lists all the possible types of questions from a said topic? In this case the said topic being co-ordinate geometry. Please assist.
Math Expert V
Joined: 02 Sep 2009
Posts: 59685

### Show Tags

Shbm wrote:
Hey Bunuel is there a directory kinda thing which lists all the possible types of questions from a said topic? In this case the said topic being co-ordinate geometry. Please assist.

Check below:

24. Coordinate Geometry

For other subjects:
ALL YOU NEED FOR QUANT ! ! !
_________________
Intern  B
Joined: 25 Dec 2017
Posts: 5

### Show Tags

Hi expert,

A line has no endpoints; it goes on forever in each direction. Two lines will intersect if they have different slopes. Two lines will not intersect (meaning they will be parallel) if they have the same slope but different y intercepts. keeping this in mind i am trying to solve this question please tell me it is right or wrong.

so both of them must be intersecting line .
slope of 1 line is 2 then slope of 2line must be negative for intersecting line and option B give us -1/2. So, B is the answer.

I am weak in coordinate geometry. I don't know it is right or wrong. Please tell me where i fall.
Director  V
Joined: 06 Jan 2015
Posts: 699
Location: India
Concentration: Operations, Finance
GPA: 3.35
WE: Information Technology (Computer Software)

### Show Tags

Bunuel 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 ?

A.$$y = \frac{1}{2} + \frac{x}{2}$$
B.$$2y = 1 - x$$
C.$$\frac{x + y}{2} = -1$$
D.$$y = \frac{x}{2} - 1$$
E.$$x = 2y + 1$$

If the equation of line 1 is $$y = 2x + 1$$

When you sub x=0, y=1 & if y=0 x=-1/2

So when $$(b, -a)$$ ==> (1, 1/2)

Gladiator59, Bunuel, generis When you sub (1, 1/2) in Eq B, it should agree right?

But it is not, in this case Why?

Where am I going wrong?
_________________
आत्मनॊ मोक्षार्थम् जगद्धिताय च

Resource: GMATPrep RCs With Solution Re: M18-24   [#permalink] 18 Mar 2019, 07:27
Display posts from previous: Sort by

# M18-24

Moderators: chetan2u, Bunuel  