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Lines a and b have different yintercepts. When line a is [#permalink]
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20 Mar 2011, 16:15
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Lines a and b have different yintercepts. When line a is reflected around the yaxis, is its reflection parallel to line b? (1) Line a is perpendicular to line b. (2) The slope of line b > 0.
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Re: If you got the last question right, check this out too. [#permalink]
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20 Mar 2011, 21:57
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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Re: If you got the last question right, check this out too. [#permalink]
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20 Mar 2011, 22:09
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! Posted from my mobile device



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Re: If you got the last question right, check this out too. [#permalink]
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21 Mar 2011, 15:34
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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Re: If you got the last question right, check this out too. [#permalink]
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21 Mar 2011, 19:55
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 Posted from my mobile device



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Re: If you got the last question right, check this out too. [#permalink]
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21 Mar 2011, 20:22
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 xcoordinate The reflection of the point (x, y) across the yaxis 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 xaxis is the point (x, y).



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Re: If you got the last question right, check this out too. [#permalink]
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21 Mar 2011, 20:35
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 Yaxis) 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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Re: If you got the last question right, check this out too. [#permalink]
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21 Mar 2011, 21:07
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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Re: If you got the last question right, check this out too. [#permalink]
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21 Mar 2011, 21:32
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.
Posted from my mobile device 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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Re: If you got the last question right, check this out too. [#permalink]
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21 Mar 2011, 21:44
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 (y3)/(x2) =1 y3 =x2 y=x+1 (1)
The equation of the reflected line (slope 1) is (y3)/[x(2)]=1 y3=1(x+2) y3 = 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, 03:40, edited 1 time in total.



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Re: If you got the last question right, check this out too. [#permalink]
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21 Mar 2011, 22:01
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 Y2/(x3) =1 Y2 =x3 Y=x1
The equation of the reflected line is Y(2)/(x3)=1 y+2=x+3 y=x+1=1x
Posted from my mobile device 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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Re: If you got the last question right, check this out too. [#permalink]
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23 Mar 2011, 04:53
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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 4564 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 ... hegraphs/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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Re: Coordinate Geometry Problem [#permalink]
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19 Jun 2011, 08:54
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guygmat wrote: Lines a and b have different yintercepts. When line a is reflected around the yaxis, 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



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Re: Lines a and b have different yintercepts. [#permalink]
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10 Nov 2012, 16:14
let the lines be a > y = mx + c b > y = nx + d
reflection of line 'a' around yaxis 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



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Re: Cant draw the graph to reach the solution [#permalink]
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28 Aug 2013, 22:39
Shibs wrote: Lines a and b have different yintercepts. When line a is reflected around the yaxis, 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: linesaandbhavedifferentyinterceptswhenlineais111182.html#p897728
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Re: Lines a and b have different yintercepts. When line a is [#permalink]
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