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For every point \((a, b)\) lying on line 1, point \((b, a)\) lies on line 2. If the equation of line 1 is \(y = 2x + 1\), what is the equation of line 2 ? A. \(y = \frac{1}{2} + \frac{x}{2}\) B. \(2y = 1  x\) C. \(\frac{x + y}{2} = 1\) D. \(y = \frac{x}{2}  1\) E. \(x = 2y + 1\)
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16 Sep 2014, 01:04
Official Solution:For every point \((a, b)\) lying on line 1, point \((b, a)\) lies on line 2. If the equation of line 1 is \(y = 2x + 1\), what is the equation of line 2 ? A. \(y = \frac{1}{2} + \frac{x}{2}\) B. \(2y = 1  x\) C. \(\frac{x + y}{2} = 1\) D. \(y = \frac{x}{2}  1\) E. \(x = 2y + 1\) Find two points on line 2 and use their coordinates to build the line's equation. Points \((0, 1)\) and \((\frac{1}{2}, 0)\) on line 1 correspond to points \((1, 0)\) and \((0, \frac{1}{2})\) on line 2. The equation of line 2 is \(\frac{y  0}{\frac{1}{2}  0} = \frac{x  1}{0  1}\) or \(2y = 1  x\). Answer: B
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Re: M1824
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17 Dec 2014, 09:51
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
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Re: M1824
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24 May 2016, 13:46
I think this can also be solved by way of negative reciprocals since the two lines are perpendicular.
From the premise it seems that is the case since every point (a,b) is (b, a) on the other line (pointing to a different direction and y and x being reversed.
Hence the answer will be the equation with slope =  1/2. (To find the negative reciprocal of a number, flip the number over (invert) and negate that value)



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Re: M1824
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13 May 2017, 09:52
Bunuel wrote: Official Solution:
For every point \((a, b)\) lying on line 1, point \((b, a)\) lies on line 2. If the equation of line 1 is \(y = 2x + 1\), what is the equation of line 2 ?
A. \(y = \frac{1}{2} + \frac{x}{2}\) B. \(2y = 1  x\) C. \(\frac{x + y}{2} = 1\) D. \(y = \frac{x}{2}  1\) E. \(x = 2y + 1\)
Find two points on line 2 and use their coordinates to build the line's equation. Points \((0, 1)\) and \((\frac{1}{2}, 0)\) on line 1 correspond to points \((1, 0)\) and \((0, \frac{1}{2})\) on line 2. The equation of line 2 is \(\frac{y  0}{\frac{1}{2}  0} = \frac{x  1}{0  1}\) or \(2y = 1  x\).
Answer: B Could you please elaborate on why \(\frac{y  0}{\frac{1}{2}  0} = \frac{x  1}{0  1}\) or \(2y = 1  x\)?



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Re: M1824
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13 May 2017, 09:57
Evgart wrote: Bunuel wrote: Official Solution:
For every point \((a, b)\) lying on line 1, point \((b, a)\) lies on line 2. If the equation of line 1 is \(y = 2x + 1\), what is the equation of line 2 ?
A. \(y = \frac{1}{2} + \frac{x}{2}\) B. \(2y = 1  x\) C. \(\frac{x + y}{2} = 1\) D. \(y = \frac{x}{2}  1\) E. \(x = 2y + 1\)
Find two points on line 2 and use their coordinates to build the line's equation. Points \((0, 1)\) and \((\frac{1}{2}, 0)\) on line 1 correspond to points \((1, 0)\) and \((0, \frac{1}{2})\) on line 2. The equation of line 2 is \(\frac{y  0}{\frac{1}{2}  0} = \frac{x  1}{0  1}\) or \(2y = 1  x\).
Answer: B Could you please elaborate on why \(\frac{y  0}{\frac{1}{2}  0} = \frac{x  1}{0  1}\) or \(2y = 1  x\)? \(\frac{y  0}{\frac{1}{2}  0} = \frac{x  1}{0  1}\); \(\frac{y }{\frac{1}{2}} = \frac{x  1}{ 1}\); \(2y=x+1\); \(2y = 1x\).
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Re: M1824
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14 May 2017, 05:59
Bunuel wrote: Evgart wrote: Bunuel wrote: Official Solution:
For every point \((a, b)\) lying on line 1, point \((b, a)\) lies on line 2. If the equation of line 1 is \(y = 2x + 1\), what is the equation of line 2 ?
A. \(y = \frac{1}{2} + \frac{x}{2}\) B. \(2y = 1  x\) C. \(\frac{x + y}{2} = 1\) D. \(y = \frac{x}{2}  1\) E. \(x = 2y + 1\)
Find two points on line 2 and use their coordinates to build the line's equation. Points \((0, 1)\) and \((\frac{1}{2}, 0)\) on line 1 correspond to points \((1, 0)\) and \((0, \frac{1}{2})\) on line 2. The equation of line 2 is \(\frac{y  0}{\frac{1}{2}  0} = \frac{x  1}{0  1}\) or \(2y = 1  x\).
Answer: B Could you please elaborate on why \(\frac{y  0}{\frac{1}{2}  0} = \frac{x  1}{0  1}\) or \(2y = 1  x\)? \(\frac{y  0}{\frac{1}{2}  0} = \frac{x  1}{0  1}\); \(\frac{y }{\frac{1}{2}} = \frac{x  1}{ 1}\); \(2y=x+1\); \(2y = 1x\). Thank you, I can understand that:) My question is more about why: \(\frac{y  0}{\frac{1}{2}  0} = \frac{x  1}{0  1}\).[/quote]



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Re: M1824
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14 May 2017, 06:05
Evgart wrote: Thank you, I can understand that:) My question is more about why:
\(\frac{y  0}{\frac{1}{2}  0} = \frac{x  1}{0  1}\).
[/quote] This is the way to find equation of a line given two points. Check coordinate geometry chapter of math book for more: https://gmatclub.com/forum/mathcoordin ... 87652.html
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Bunuel : where am I going wrong? Line equation passing through two points (1,0) and (0,1/2)  line 2. Equation of line => yy1/xx1 = y1y2/x1x2 y0/x1 = (01/2)/(10) y/x1 = 1/2 y = 1/2(x1) y= 1/2x + 1/2



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Re: M1824
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04 Jul 2017, 08:01
ManSab wrote: Bunuel : where am I going wrong? Line equation passing through two points (1,0) and (0,1/2)  line 2. Equation of line => yy1/xx1 = y1y2/x1x2 y0/x1 = (01/2)/(10) y/x1 = 1/2 y = 1/2(x1) y= 1/2x + 1 I was not able to decipher your way above because of poor formatting. Anyway below is correct way: \(\frac{y  0}{\frac{1}{2}  0} = \frac{x  1}{0  1}\); \(\frac{y }{\frac{1}{2}} = \frac{x  1}{ 1}\); \(2y=x+1\); \(2y = 1x\).
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Re: M1824
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04 Jul 2017, 08:58
@Buneul  not able to figure out your steps, hence asking your review for my steps. This time with better format. Equation of line formula = \(\frac{y−y1}{xx1}\) =\(\frac{y1y2}{x1x2}\) Using your example line points (1,0) and (0,1/2)
\(\frac{y0}{x1}\) =\(\frac{0  \frac{1}{2}}{10}\)
\(\frac{y}{x1}\)=\(\frac{1}{2}\)
y =\(\frac{1}{2}(x1)\) y=\(\frac{1}{2}x+\frac{1}{2}\)
Thanks for your inputs!



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Re: M1824
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04 Jul 2017, 09:03
ManSab wrote: @Buneul  not able to figure out your steps, hence asking your review for my steps. This time with better format. Equation of line formula = \(\frac{y−y1}{xx1}\) =\(\frac{y1y2}{x1x2}\) Using your example line points (1,0) and (0,1/2)
\(\frac{y0}{x1}\) =\(\frac{0  \frac{1}{2}}{10}\)
\(\frac{y}{x1}\)=\(\frac{1}{2}\)
y =\(\frac{1}{2}(x1)\) y=\(\frac{1}{2}x+\frac{1}{2}\)
Thanks for your inputs! No mistake there. Multiply by 2: \(2y = x + 1\) or \(2y = 1  x\).
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Re: M1824
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04 Jul 2017, 16:57
Thank you Bunuel.!



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My approach:
Let a = 2. Therefore x = 2. Substitute in L1's eqn and solve for y. Thus, y = 5 (i.e. b = 5). This means there is a point on L2 with the coordinates (5, 2). Substitute either of these points in the answer choices and identify the equation that, when solved, gives you the other point.



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Re: M1824
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11 May 2018, 09:24
Hi Bunuel, Can you explain how did you get (0,1) on line 1 and (1/2,0) on line 2.



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11 May 2018, 09:33
hero_with_1000_faces wrote: Hi Bunuel, Can you explain how did you get (0,1) on line 1 and (1/2,0) on line 2. The equation of line 1 is y = 2x + 1. If x = 0, then y = 2*0 + 1 = 1. So, point (0, 1) is on line y = 2x + 1. The stem says: for every point (a,b) lying on line 1, point (b,−a) lies on line 2. Point (a = 0, b = 1) is on line 1, so point (b = 1, a = 0) is on line 2. If y = 0, then 0 = 2x + 1 > x = 1/2. So, point (1/2, 0) is on line y = 2x + 1. The stem says: for every point (a,b) lying on line 1, point (b,−a) lies on line 2. Point (a= 1/2, b = 0) is on line 1, then point (b = 0, a = 1/2) is on line 2.
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Re: M1824
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05 Nov 2018, 22:49
Hey Bunuel is there a directory kinda thing which lists all the possible types of questions from a said topic? In this case the said topic being coordinate geometry. Please assist.
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Re: M1824
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05 Nov 2018, 23:40
Shbm wrote: Hey Bunuel is there a directory kinda thing which lists all the possible types of questions from a said topic? In this case the said topic being coordinate geometry. Please assist. Check below: 24. Coordinate Geometry For other subjects: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative Megathread
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Re: M1824
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06 Dec 2018, 12:03
Hi expert,
A line has no endpoints; it goes on forever in each direction. Two lines will intersect if they have different slopes. Two lines will not intersect (meaning they will be parallel) if they have the same slope but different y intercepts. keeping this in mind i am trying to solve this question please tell me it is right or wrong.
(a,b) lie on 1st quadrant (b,a) lie on 2nd quadrant so both of them must be intersecting line . slope of 1 line is 2 then slope of 2line must be negative for intersecting line and option B give us 1/2. So, B is the answer.
I am weak in coordinate geometry. I don't know it is right or wrong. Please tell me where i fall.



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Re: M1824
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18 Mar 2019, 07:27
Bunuel wrote: For every point \((a, b)\) lying on line 1, point \((b, a)\) lies on line 2. If the equation of line 1 is \(y = 2x + 1\), what is the equation of line 2 ?
A.\(y = \frac{1}{2} + \frac{x}{2}\) B.\(2y = 1  x\) C.\(\frac{x + y}{2} = 1\) D.\(y = \frac{x}{2}  1\) E.\(x = 2y + 1\) If the equation of line 1 is \(y = 2x + 1\) When you sub x=0, y=1 & if y=0 x=1/2 So when \((b, a)\) ==> (1, 1/2) Gladiator59, Bunuel, generis When you sub (1, 1/2) in Eq B, it should agree right? But it is not, in this case Why? Where am I going wrong?
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