Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
16 Jul 2009, 20:58
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0%
(00:00)
wrong
based on 0
sessions
History
Date
Time
Result
Not Attempted Yet
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?
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?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
18 Jul 2009, 09:42
Kudos
Bookmarks
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?
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?
18 Jul 2009, 12:43
Kudos
Bookmarks
Economist
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.
