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 2308:00 PM PDT
-09:00 PM PDT
Video explanations + diagnosis of 10 weakest areas + 150+ short videos + study plan!
Apr 2301:30 AM EDT
-02:30 AM EDT
Struggling with GMAT Verbal as a non-native speaker? Harsh improved his score from 595 to 695 in just 45 days—and scored a 99 %ile in Verbal (V88)! Learn how smart strategy, clarity, and guided prep helped him gain 100 points.
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
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.
30 Jan 2011, 13:55
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
65% (02:35) correct
35%
(02:25)
wrong
based on 325
sessions
History
Date
Time
Result
Not Attempted Yet
For every point \((a, b)\) on Line 1, there is a corresponding point \((b, -a)\) on Line 2. Given that the equation of Line 1 is \(y = 2x + 1\), what is the equation of Line 2?
A. \(y = \frac{x}{2} + \frac{1}{2} \)
B. \(y = -\frac{x}{2} + \frac{1}{2}\)
C. \(y = -\frac{x}{2} - \frac{1}{2}\)
D. \(y = \frac{x}{2} - \frac{1}{2}\)
E. \(y = -\frac{x}{2} +1\)
A. \(y = \frac{x}{2} + \frac{1}{2} \)
B. \(y = -\frac{x}{2} + \frac{1}{2}\)
C. \(y = -\frac{x}{2} - \frac{1}{2}\)
D. \(y = \frac{x}{2} - \frac{1}{2}\)
E. \(y = -\frac{x}{2} +1\)
Experience GMAT Club Test Questions
Yes, you've landed on a GMAT Club Tests question
Craving more? Unlock our full suite of GMAT Club Tests here
Want to experience more? Get a taste of our tests with our free trial today
Rise to the challenge with GMAT Club Tests. Happy practicing!
12 Jan 2015, 12:49
Kudos
Bookmarks
For every point \((a, b)\) on Line 1, there is a corresponding point \((b, -a)\) on Line 2. Given that the equation of Line 1 is \(y = 2x + 1\), what is the equation of Line 2?
A. \(y = \frac{x}{2} + \frac{1}{2} \)
B. \(y = -\frac{x}{2} + \frac{1}{2}\)
C. \(y = -\frac{x}{2} - \frac{1}{2}\)
D. \(y = \frac{x}{2} - \frac{1}{2}\)
E. \(y = -\frac{x}{2} +1\)
To find the equation of Line 2, we can first identify two points on Line 1 and then determine their corresponding points on Line 2. Points \((0, 1)\) and \((1, 3)\) lie on Line 1, so their corresponding points on Line 2 are \((1, 0)\) and \((3, -1)\), respectively.
Using the form \(y = mx + b\) for the equation of Line 2, we can employ the two points we found to create a system of equations to solve for the slope \(m\) and the y-intercept \(b\). Plugging the coordinates of the points into the equation, we obtain:
\(0 = m(1) + b\)
\(-1 = m(3) + b\)
Upon solving this system of equations, we find that \(m = -\frac{1}{2}\) and \(b = \frac{1}{2}\).
Consequently, the equation of Line 2 is \(y = -\frac{x}{2} + \frac{1}{2}\).
Answer: B
A. \(y = \frac{x}{2} + \frac{1}{2} \)
B. \(y = -\frac{x}{2} + \frac{1}{2}\)
C. \(y = -\frac{x}{2} - \frac{1}{2}\)
D. \(y = \frac{x}{2} - \frac{1}{2}\)
E. \(y = -\frac{x}{2} +1\)
To find the equation of Line 2, we can first identify two points on Line 1 and then determine their corresponding points on Line 2. Points \((0, 1)\) and \((1, 3)\) lie on Line 1, so their corresponding points on Line 2 are \((1, 0)\) and \((3, -1)\), respectively.
Using the form \(y = mx + b\) for the equation of Line 2, we can employ the two points we found to create a system of equations to solve for the slope \(m\) and the y-intercept \(b\). Plugging the coordinates of the points into the equation, we obtain:
\(0 = m(1) + b\)
\(-1 = m(3) + b\)
Upon solving this system of equations, we find that \(m = -\frac{1}{2}\) and \(b = \frac{1}{2}\).
Consequently, the equation of Line 2 is \(y = -\frac{x}{2} + \frac{1}{2}\).
Answer: B
General Discussion
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
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
I hope it makes sense now.
