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Lines a and b have different y-intercepts. When line a is

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Lines a and b have different y-intercepts. When line a is [#permalink]

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20 Mar 2011, 17:15
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Lines a and b have different y-intercepts. When line a is reflected around the y-axis, is its reflection parallel to line b?

(1) Line a is perpendicular to line b.

(2) The slope of line b > 0.
[Reveal] Spoiler: OA

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20 Mar 2011, 22:57
S1 insufficient
Y axis has no reflection about y axis. Assume line a= y axis. The answer is NO

If line a has slope -1. Slop of b is 1. Reflection of line a about y axis is parallel to line b. The answer is YES

S2 insufficient.
Slope of line a is unknown

1) + 2) sufficient
Since b>0 the line b cannot be x axis. Line a cannot be y axis - a is perpendicular to b. That means when a has reflection about y axis it is parallel to b. The answer to the question is YES

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20 Mar 2011, 23:09
But my initial guess was A alone sine we know that line a and line b have known and "different y intercepts". This precludes a from being the y axis . So I have two answers a or c. My bet 50/50 on both tough one!

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21 Mar 2011, 16:34
But according to the 'bagrettin' explanation to my previous question:
a --- y = mx + b
b --- y = (-1/m)x + b
Reflection of line a --- x = ym + c ( shouldn't we flip x and y values to find reflection?)
Y = (1/m) x - ( c/ m )
So now slope of line b is (-1/m) & slope of reflected line is (1/m)
It looks like they will never be parallel because if one is + ve another is - ve.
Help me ? I don't know whether we should proceed this way or not.
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21 Mar 2011, 20:55
Hello bhandariavi
x coordinate flips sign due to reflection about y axis. Draw two perpendicular lines in any quadrant and you will see that reflection of line a (whose slope -1) Reflected line will have slope = 1 is parallel to line b (slope 1) - ( knowing that line a is perpendicular to b) Visualize the problem that is easier than algebra or you can request bagrettin for algebraic solution. Cheers

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21 Mar 2011, 21:22
bhandariavi wrote:
But according to the 'bagrettin' explanation to my previous question:
a --- y = mx + b
b --- y = (-1/m)x + b
Reflection of line a --- x = ym + c ( shouldn't we flip x and y values to find reflection?)
Y = (1/m) x - ( c/ m )
So now slope of line b is (-1/m) & slope of reflected line is (1/m)
It looks like they will never be parallel because if one is + ve another is - ve.
Help me ? I don't know whether we should proceed this way or not.

bhandariavi,
For reflection on y- axis you don't have to flip x and y values .Just negate x-coordinate
The reflection of the point (x, y) across the y-axis is the point (-x, y).
The reflection of the point (x, y) across the line y = x is the point (y, x).
The reflection of the point (x, y) across the line y = -x is the point (-y, -x).
The reflection of the point (x, y) across the x-axis is the point (x, -y).
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21 Mar 2011, 21:35
subhashghosh wrote:
The answer is E, a graph can be drawn to visualize this.

a --- y = mx + b
b --- y = (m1)x + b
Reflection of line a --- y = -mx + c (Reflection over Y-axis)
we got to find if (-m)= (m1)

Statement A : m *m1=-1..
m=-1/m1
M1=-1 AND M=1, 1=1 no
M1=1 AND M=1, -1=1 yes
So InSufficient...

Statement B b>0... Insufficient...

So It should be E....
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21 Mar 2011, 22:07
Hello onell
Due to reflection about y axis Not only the x flips sign but the slope of the line "also" flips sign. I believe your equation did not account for slope change. Pls correct me if this not true.

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21 Mar 2011, 22:32
gmat1220 wrote:
Hello onell
Due to reflection about y axis Not only the x flips sign but the slope of the line "also" flips sign. I believe your equation did not account for slope change. Pls correct me if this not true.

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If it flips the sign of x coordinate and changes the sign of a slope : You will get a original line
Consider a line passes through (x ,y)
y = mx + b

Upon reflection it passes through (-x ,y) and changes slope to -m (As you have written)
y = -m (-x) + c
y=mx +c (Equation of a original line)
Am I missing sth ?
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21 Mar 2011, 22:44
Hello onell
Strange! Let me try will coordinates. Let's say line a (slope 1) passes through (2,3) and upon reflection passes through (-2,3) slope=-1
Equation of line a (slope 1) is
(y-3)/(x-2) =1
y-3 =x-2
y=x+1 -----------(1)

The equation of the reflected line (slope -1) is
(y-3)/[x-(-2)]=-1
y-3=-1(x+2)
y-3 = -x - 2
y = -x + 1 ----------(2)

Two different equations. The slope of the line b which is perpendicular to a is -1. The slope of the reflected line is also -1. Hence E cannot be the answer.

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Last edited by gmat1220 on 22 Mar 2011, 04:40, edited 1 time in total.
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21 Mar 2011, 23:01
gmat1220 wrote:
Hello onell
Strange! Let me try will coordinates. Let's say line a (slope 1) passes through (2,3) and upon reflection passes through (-2,3) slope=-1
Equation of line a is
Y-2/(x-3) =1
Y-2 =x-3
Y=x-1

The equation of the reflected line is
Y-(-2)/(x-3)=-1
y+2=-x+3
y=-x+1=1-x

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Strange Indeed. However if you substitute (2,3) and slope 1 and(-2,3) and slope -1 for reflected line in equation y=mx+c . You get the same equation for both the line..
Bunuel, Please help....
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23 Mar 2011, 05:53
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Upon request, I am providing the solution of the question above:
I will provide a graphical approach since that is what I favor always but later will give an algebraic approach too...

The two diagrams below illustrate the case where we take both statements together. In one case, reflection of a is not parallel to b and in the other reflection of a is parallel to b. Hence even with both statements we cannot say whether reflection of a is parallel to b. Answer (E).

Attachment:

Ques2.jpg [ 23.2 KiB | Viewed 3558 times ]

If a and b make 45 degrees angle with the y axis (as shown, technically I will not say that they are both making 45 degrees angle with y axis but let's not worry about it here), when a is reflected along y axis, its angle with y axis is still 45. In this case a and b are parallel.

Algebraic approach:
A line is defined by 2 things - its slope and y intercept. When we reflect a line along the y axis, its slope flips sign but y intercept remains unchanged.
For more on this: http://www.veritasprep.com/blog/2010/12 ... he-graphs/
Now,
a -> y = mx + c
b -> y = nx + d
Reflected a -> y = -mx + c
Ques: Is -m = n? (Parallel lines have the same slope.)

Stmnt 1: mn = -1
m = -1/n.
If m = 1 and n = -1, -m is equal to n
If m = -1 and n = 1, -m is equal to n
If m = -1/2 and n = 2, -m is not equal to n
Not sufficient.

Stmnt 2: n>0
If m = -1 and n = 1, -m is equal to n
If m = -1/2 and n = 2, -m is not equal to n
Not sufficient.

Taking both together,
If m = -1 and n = 1, -m is equal to n
If m = -1/2 and n = 2, -m is not equal to n
Not sufficient. Answer (E).
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 18 Jun 2011 Posts: 58 Followers: 1 Kudos [?]: 105 [0], given: 0 Lines a and b have different y-intercepts. When line a is [#permalink] Show Tags 19 Jun 2011, 08:04 1 This post was BOOKMARKED Lines a and b have different y-intercepts. When line a is reflected around the y-axis, is its reflection parallel to line b? (1) Line a is perpendicular to line b. (2) The slope of line b > 0. Current Student Joined: 26 May 2005 Posts: 565 Followers: 18 Kudos [?]: 194 [1] , given: 13 Re: Coordinate Geometry Problem [#permalink] Show Tags 19 Jun 2011, 09:54 1 This post received KUDOS guygmat wrote: Lines a and b have different y-intercepts. When line a is reflected around the y-axis, is its reflection parallel to line b? (1) Line a is perpendicular to line b. (2) The slope of line b > 0. line a = y =mx+b line b = y = nx+c now in case a line reflects, its slope changes sign and Y intercept remains same so relected a = y= -mx+b now the question asks is -m =n ( parallel lines will have same slope) st1. -m*n = -1 m = 1 and n = 1 ... satisfies our condition m = 1/3 and n = 3 ; -m*n = -1 but -m is not equal to n ... hence in sufficient st2: n>0 same as above... insufficient 1&2 m = 1 n =1 ok m = 1/3 and n = 3 not ok hence E Intern Status: ISB 14...:) Joined: 26 May 2012 Posts: 30 Location: India Concentration: Strategy Schools: ISB '14 (A) GMAT 1: 750 Q51 V39 GPA: 3.62 WE: Engineering (Energy and Utilities) Followers: 1 Kudos [?]: 31 [0], given: 11 Re: Lines a and b have different y-intercepts. [#permalink] Show Tags 10 Nov 2012, 17:14 let the lines be a ---> y = mx + c b ---> y = nx + d reflection of line 'a' around y-axis would be y = - mx + c from the first statement, mn = - 1 but we cannot decide whether they are parellel as '- m' can be or cannot be equal to 'n'. from the second statement, n>0 this signifies nothing regarding the relationship between slopes of the two lines, hence not sufficient Even after combining both the statements, the data is insufficient For example, if m = -1; n = 1 it satisfies both the statements and the lines are parellel and for m = -0.5; n = 2 it satisfies both the equations and the lines are not parellel. Ans. E Current Student Joined: 05 Aug 2012 Posts: 42 Concentration: Technology, General Management GMAT Date: 09-16-2013 GPA: 3.5 WE: Analyst (Consulting) Followers: 0 Kudos [?]: 23 [0], given: 11 Re: Cant draw the graph to reach the solution [#permalink] Show Tags 28 Aug 2013, 20:17 Shibs wrote: Lines a and b have different y-intercepts. When line a is reflected around the y-axis, is its reflection parallel to line b? (1) Line a is perpendicular to line b. (2) The slope of line b > 0. IMO A. You can get it using the laws of reflection. Correct me if wrong. _________________ Regards, Suyash I want to live in a world where emails are short, love letters are brave, and every "Thank you" note is scribbled by hand. GO GREEN Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6829 Location: Pune, India Followers: 1919 Kudos [?]: 11928 [1] , given: 221 Re: Cant draw the graph to reach the solution [#permalink] Show Tags 28 Aug 2013, 23:39 1 This post received KUDOS Expert's post Shibs wrote: Lines a and b have different y-intercepts. When line a is reflected around the y-axis, is its reflection parallel to line b? (1) Line a is perpendicular to line b. (2) The slope of line b > 0. The answer here is (E). I have provided the graphical and algebraic approaches here: lines-a-and-b-have-different-y-intercepts-when-line-a-is-111182.html#p897728 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Cant draw the graph to reach the solution [#permalink]

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29 Aug 2013, 01:56
Shibs wrote:
Lines a and b have different y-intercepts. When line a is reflected around the y-axis, is its reflection parallel to line b?

(1) Line a is perpendicular to line b.

(2) The slope of line b > 0.

Merging similar topics.

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Re: Lines a and b have different y-intercepts. When line a is [#permalink]

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Lines a and b have different y-intercepts. When line a is

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