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.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
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.
31 Mar 2012, 02:54
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
75% (01:41) correct
25%
(02:21)
wrong
based on 1118
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
Capture.GIF [ 3.28 KiB | Viewed 67409 times ]
A. (-4, -1)
B. (-1, 4)
C. (4, -1)
D. (1, -4)
E. (4, 1)
31 Mar 2012, 04:26
Kudos
Bookmarks
enigma123
Since the line y=x is the perpendicular bisector of segment AB, then the point B is the mirror reflection of point A around the line y=x, so its coordinates are (4, 1). The same way, since the x-axis is the perpendicular bisector of segment BC then the point C is the mirror reflection of point B around the x-axis, so its coordinates are (4, -1).
Answer: C.
The question becomes much easier if you just draw rough sketch of the diagram:
Attachment:
graph.png [ 12.57 KiB | Viewed 67091 times ]
Answer: C.
Similar questions to practice:
in-the-xy-coordinate-plane-is-point-r-equidistant-from-143502.html
in-the-rectangular-coordinate-system-the-line-y-x-is-the-132646.html
the-coordinates-of-points-a-and-c-are-0-3-and-127769.html
the-line-represented-by-the-equation-y-4-2x-is-the-127770.html
if-point-a-coordinates-are-7-3-point-b-coordinates-a-141972.html
in-the-rectangular-coordinate-system-the-line-y-x-is-the-88473.html
in-the-rectangular-coordinate-system-above-the-line-y-x-144774.html
the-line-represented-by-the-equation-y-4-2x-is-the-perpendi-87573.html
Hope it helps.
01 Aug 2012, 06:56
Kudos
Bookmarks
teal
The mirror image of \((x,y)\) around the Y-axis is \((-x,y)\).
For the second question:
Assume we have a point P(a,b) and we want to find its mirror image around the line \(y = -x\).
Let's denote the point we seek by Q(A,B). See the attached drawing.
The equation of the line passing through P and perpendicular to the line \(y = -x\) is \(y - b = x - a\), or \(y = x + b - a\).
Since Q is also on this line, we have \(B = A + b - a\), from which \(A - B = a - b\).
The middle point of the line segment PQ (denoted by M) is also on the line \(y = -x\), therefore \(\frac{a+A}{2}=-\frac{b+B}{2}\), or \(A + B = -a - b\).
Solving for A and B, we find that \(A = -b, B =-a\).
Therefore, the mirror image of \((x,y)\) around the line \(y = -x\) is \((-y, -x)\).
Attachments
MirrorImage.jpg [ 18.1 KiB | Viewed 64681 times ]
