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16 Sep 2014, 01:04
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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. \(y = -\frac{x}{2} +1\)
16 Sep 2014, 01:04
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Official Solution:
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 \(y = -\frac{x}{2} + \frac{1}{2}\).
Note that while Geometry is not tested on GMAT Focus, Coordinate Geometry is tested under the Functions and Graphing sections found in the Official Guide for GMAT Focus Edition.
Answer: B
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 \(y = -\frac{x}{2} + \frac{1}{2}\).
Note that while Geometry is not tested on GMAT Focus, Coordinate Geometry is tested under the Functions and Graphing sections found in the Official Guide for GMAT Focus Edition.
Answer: B
General Discussion
17 Dec 2014, 09:51
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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
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
