It is currently 17 Mar 2018, 15:16

### 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

# OG 11-199 > Reflection of a point thru line

Author Message
Director
Joined: 01 Apr 2008
Posts: 848
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
OG 11-199 > Reflection of a point thru line [#permalink]

### Show Tags

16 Jul 2009, 20:58
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Only for those who have OG 11 I dont want to create a diagram. This post is to discuss about a general concept.

The explanation says something about "Reflection of point A through line y=x". Since B is the reflection co-ordinates of B can be derived by swapping x and y coordinates of A.

This looks very logical however, I think such reflection rule can be applied only to X-axis, Y-axis or lines that make a 45 degrees angle with the axes ( of course the co-ordinates and signs can change based on the quadrant ).

Is my understanding correct? or is this reflection rule generic for a line with any slope?
Manager
Joined: 03 Jul 2009
Posts: 103
Location: Brazil
Re: OG 11-199 > Reflection of a point thru line [#permalink]

### Show Tags

18 Jul 2009, 09:42
This reflection rule is generic for a line with any slope.

If you know how to work well with distances in the coordinate plane, you will not have to much trouble.

You must calculate the distance between the given point to the line, and then with that same distance, you will find the reflection point.

Did I help?
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1346
Re: OG 11-199 > Reflection of a point thru line [#permalink]

### Show Tags

18 Jul 2009, 12:43
Economist wrote:
Is my understanding correct? or is this reflection rule generic for a line with any slope?

It's certainly not generic - finding points of reflection is much more difficult if you aren't dealing with a simple line like y = x, y = -x, y = 0 or x = 0. It's not something you'll likely ever need to do on a real GMAT. Just to see how much work is involved, let's take a pretty simple line and point:

What are the co-ordinates of the point obtained when (4,0) is reflected in the line y = 2x?

Of course, depending on the answer choices, all you might need to do here is draw the picture, make a good estimate, and if only one answer is close, you're done. That can be a very good approach on some coordinate geometry questions, and will often at least let you rule out a few wrong answer choices. We can work out the exact co-ordinates:

Draw a line from the point (4,0) through the line y = 2x, which is perpendicular to y = 2x. We want to find the point on this line on the opposite side of y = 2x which is the same distance to y = 2x as (4,0). This new line is perpendicular to y = 2x, so its slope is -1/2. Plugging (4,0) into y = (-1/2)x + b, we find that b = 2. Now, y = (-1/2)x + 2 intersects y = 2x at the point (4/5, 8/5); let's call this point P. To get from (4,0) to (4/5, 8/5), we go up by 8/5, and to the left by 16/5. If we do those steps again but starting from the point (4/5, 8/5), we'll find a point on the line y = (-1/2)x + 2 which is as far from P as (4,0) is. So our reflection point is (4/5 - 16/5, 8/5 + 8/5) = (-12/5, 16/5). Or you could use the distance formula to complete this last step, though that leads to a more complicated calculation.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Manager
Joined: 03 Jul 2009
Posts: 103
Location: Brazil
Re: OG 11-199 > Reflection of a point thru line [#permalink]

### Show Tags

18 Jul 2009, 13:07
Oh sorry, When I copy you "This reflection rule is generic for a line with any slope.", and did not notice that you were referring to the fact of just swapping x and y coordinates.

I meant that the principle is the same, but yes, as IanStewart said: "It's not something you'll likely ever need to do on a real GMAT.", because of the work to do that.
Manager
Joined: 27 Jun 2008
Posts: 144
Re: OG 11-199 > Reflection of a point thru line [#permalink]

### Show Tags

22 Jul 2009, 00:50
Concur with Ian, Diagramming would be easier, wasteful to spend so much time calculating the coordinates.
Manager
Joined: 14 Nov 2008
Posts: 189
Schools: Stanford...Wait, I will come!!!
Re: OG 11-199 > Reflection of a point thru line [#permalink]

### Show Tags

23 Jul 2009, 05:02
IanStewart wrote:
Economist wrote:
Is my understanding correct? or is this reflection rule generic for a line with any slope?

It's certainly not generic - finding points of reflection is much more difficult if you aren't dealing with a simple line like y = x, y = -x, y = 0 or x = 0. It's not something you'll likely ever need to do on a real GMAT. Just to see how much work is involved, let's take a pretty simple line and point:

What are the co-ordinates of the point obtained when (4,0) is reflected in the line y = 2x?

Of course, depending on the answer choices, all you might need to do here is draw the picture, make a good estimate, and if only one answer is close, you're done. That can be a very good approach on some coordinate geometry questions, and will often at least let you rule out a few wrong answer choices. We can work out the exact co-ordinates:

Draw a line from the point (4,0) through the line y = 2x, which is perpendicular to y = 2x. We want to find the point on this line on the opposite side of y = 2x which is the same distance to y = 2x as (4,0). This new line is perpendicular to y = 2x, so its slope is -1/2. Plugging (4,0) into y = (-1/2)x + b, we find that b = 2. Now, y = (-1/2)x + 2 intersects y = 2x at the point (4/5, 8/5); let's call this point P. To get from (4,0) to (4/5, 8/5), we go up by 8/5, and to the left by 16/5. If we do those steps again but starting from the point (4/5, 8/5), we'll find a point on the line y = (-1/2)x + 2 which is as far from P as (4,0) is. So our reflection point is (4/5 - 16/5, 8/5 + 8/5) = (-12/5, 16/5). Or you could use the distance formula to complete this last step, though that leads to a more complicated calculation.

Great Post. Thanks.
Re: OG 11-199 > Reflection of a point thru line   [#permalink] 23 Jul 2009, 05:02
Display posts from previous: Sort by