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S1) insufficient. Slopes are unknown S2) insufficient Consider line m to be inclined 60 deg to x axis and line p to be inclined 30 deg to x axis. The reflection of line p about y=x is parallel to line m. The answer to the question is yes. Consider line m and line p to be parallel to x axis. The reflection of line p about x=y is not prallel to the line m. The answer to the question is no

1) + 2) sufficient The slopes of the lines are known m=3 and p=1. we can know the angle each line makes with the x axis. And know for sure whether the reflection of line p is symmetrical about the x=y. This can be definitely answered. Sufficient. Hence C

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Last edited by gmat1220 on 21 Mar 2011, 00:12, edited 1 time in total.

Any monotonous function f(x) which is reflected over the line y=x becomes the function g(x) which is the inverse to f(x). This means f(g(y))=y for all y from the domain and g(f(x))=x for all x from the domain. In case of linear function it is very easy to find the inserve just by changing y to x and x to y. In our case the inverse x=py+q y=x/p-q/p

If this reflection is paralle to the line y=mx+n, then the slope of two lines are equal, therefore m=1/p. So, finally the question asks whether m=1/p.

(1) says m=p+2, which means that m=1/p only for some particular values of p, so we could not conclude whether the lines are parallel (2) says m=3p, which means that m=1/p only for some particular values of p, so we could not conclude whether the lines are parallel

(1) and (2) gives us the system m=p+2 m=3p

from which p+2=3p p=1 and m=2+1=3

As we could see, m is not equal to 1/p, so these lines are not parallel. Only both statements together are sufficient for the answer.

The answer is C.
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In S1. Just to be more precise, let say that correspondence between slopes p and p+2 are different for the different values of p, therefore the answer on the question will be different too (for exact equation, linking inverse function see my post)

In S2 your example with 60 and 30 degrees corresponds to m=3p. The line which is inclined 60 degrees to x axis has the slope of 3^0.5 (tg 60), while the line with 30 degrees to x axis has the slope 1/3^0.5 (tg 30), so m=3p. Again since y=x is inclined at 45 degrees to Ox, this lines are symmetrical to the y=x.

If both lines have the slope of 0, then m=3p, while the reflection will be vertical line with the slope equal to infinity.
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The explanation is indeed splendid I was searching for some more on equations of line when I found this link on tutorvista which was a great help i think you should also check it out.

Re: A line with the equation y = px + q is reflected over the li [#permalink]

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02 Oct 2012, 19:13

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A line with the equation y = px + q is reflected over the line y = x. Is the reflection of this line parallel to the line y = mx + n?

(1) m = p + 2

(2) m = 3p

When a line is reflected around y=x then the equation of the reflected line is got by swapping the x and y of the original equation. So, equation of the reflection of the line y=px + q will be x = py + q

So, for this line to be parallel to y = mx + n we just need to make sure that the slope of the two lines are equal i.e. 1/p = m

STAT1 m = p + 2 But we cannot say whether m=1/p will be true or not. Proof is below. Trying m =1/p we get 1/p = p + 2 p^2 + 2p - 1 =0 p = { -2 +- sqrt(4 + 4) }/2 = { -2 + 4sqrt2 } /2 But there can be many other values of p for which m = p+2 will be true but m = 1/p will not be true. So, INSUFFICIENT

STAT2 m = 3p But we cannot say whether m=1/p will be true or not. Proof is below. Trying m = 1/p 1/p = 3p => p^2 = 1/3 But there can be many other values of p for which m = p+2 will be true but m = 1/p will not be true. So, INSUFFICIENT

Taking both together we have m = p+2 and m=3p => 3p = p+2 => p = 1

So, m = p+2 = 3 And 1/p = 1 Clearly m is not equal to 1/p. So, lines are not parallel.

So, Answer will be C. Hope it helps!
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Re: A line with the equation y = px + q is reflected over the li [#permalink]

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02 Oct 2012, 20:33

@nktdotgupta - Thx for the explanation but i've another question on this. What if - 1. A line with the equation y = px + q is reflected over the line y = 2x, How shall we approach this question? 2. A line with the equation y = px + q is reflected over the line y = 5, How shall we approach this question?
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Re: A line with the equation y = px + q is reflected over the li [#permalink]

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02 Oct 2012, 21:40

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Hi,

To find the equation of the reflected line just keep in mind the following things:--

1. Each and every point on the reflected line has the same distance from the line on which the reflection happened as the corresponding point on the original line. 2. The line on which reflection happens is the perpendicular bisector of the line joining a point on the original line and the corresponding point on the reflected line.

So, to find the equation of the reflected line you need to find two points. One point you can get by finding the point of intersection of the original line and the line on which the reflection happened. To find the other point you can take any point on the original line and drop a perpendicular to the line on which reflection has to happen. And find the point of the intersection of the line on which the reflection has to happen and the perpendicular drawn just now. Now, the point of intersection which we got just now is the mid point of the point on the original line and the point on the reflected line. So, we can find the point on the reflected line too.

Will give an example : ( Figure attached ) Suppose we are given the equation of line l and we have to find the reflection of l on line m line n is the reflected line whose equation we need to find. We can find one point (O) by solving for lines l and m Now to find one more point on line n (C(x2,y2)): Take a point A(x,y) on line l and draw a perpendicular to line m (point B) Now , we know the slope of line m so we can find the slope of any line perpendicular to line m and we have one point (A) through which that perpendicular is passing so we can find equation of that perpendicular and hence we can find point B (x1, y1) Since, B is the midpoint of AC (since the line on which reflection happens is the perpendicular bisector of any two corresponding lines on the original and the reflected line) So, x1 = (x+ x2) /2 and y1 = (y + y2)/2 So, we can find x2,y2 and hence point C Once we have point C then we can find equation of line n as we have two point O and C

Now coming to your doubt: What if - 1. A line with the equation y = px + q is reflected over the line y = 2x, How shall we approach this question? 2. A line with the equation y = px + q is reflected over the line y = 5, How shall we approach this question?

For the 1st one i will go with the approach mentioned above. For the second one since the line is y=5. So the X co-ordinate of the corresponding point on the reflected line will remain same but for finding the y co-ordinate i will follow the procedure mentioned above.

Re: A line with the equation y = px + q is reflected over the li [#permalink]

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28 Nov 2013, 09:22

nktdotgupta wrote:

A line with the equation y = px + q is reflected over the line y = x. Is the reflection of this line parallel to the line y = mx + n?

(1) m = p + 2

(2) m = 3p

When a line is reflected around y=x then the equation of the reflected line is got by swapping the x and y of the original equation. So, equation of the reflection of the line y=px + q will be x = py + q

So, for this line to be parallel to y = mx + n we just need to make sure that the slope of the two lines are equal i.e. 1/p = m

STAT1 m = p + 2 But we cannot say whether m=1/p will be true or not. Proof is below. Trying m =1/p we get 1/p = p + 2 p^2 + 2p - 1 =0 p = { -2 +- sqrt(4 + 4) }/2 = { -2 + 4sqrt2 } /2 But there can be many other values of p for which m = p+2 will be true but m = 1/p will not be true. So, INSUFFICIENT

STAT2 m = 3p But we cannot say whether m=1/p will be true or not. Proof is below. Trying m = 1/p 1/p = 3p => p^2 = 1/3 But there can be many other values of p for which m = p+2 will be true but m = 1/p will not be true. So, INSUFFICIENT

Taking both together we have m = p+2 and m=3p => 3p = p+2 => p = 1

So, m = p+2 = 3 And 1/p = 1 Clearly m is not equal to 1/p. So, lines are not parallel.

So, Answer will be C. Hope it helps!

You say that for two lines to be parallel their product should be 1. I don't get it, I thought that for two lines to be parallel they had to have the same slope. So in this case m = p

Re: A line with the equation y = px + q is reflected over the li [#permalink]

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29 Nov 2013, 03:56

jlgdr wrote:

nktdotgupta wrote:

A line with the equation y = px + q is reflected over the line y = x. Is the reflection of this line parallel to the line y = mx + n?

(1) m = p + 2

(2) m = 3p

When a line is reflected around y=x then the equation of the reflected line is got by swapping the x and y of the original equation. So, equation of the reflection of the line y=px + q will be x = py + q

So, for this line to be parallel to y = mx + n we just need to make sure that the slope of the two lines are equal i.e. 1/p = m

STAT1 m = p + 2 But we cannot say whether m=1/p will be true or not. Proof is below. Trying m =1/p we get 1/p = p + 2 p^2 + 2p - 1 =0 p = { -2 +- sqrt(4 + 4) }/2 = { -2 + 4sqrt2 } /2 But there can be many other values of p for which m = p+2 will be true but m = 1/p will not be true. So, INSUFFICIENT

STAT2 m = 3p But we cannot say whether m=1/p will be true or not. Proof is below. Trying m = 1/p 1/p = 3p => p^2 = 1/3 But there can be many other values of p for which m = p+2 will be true but m = 1/p will not be true. So, INSUFFICIENT

Taking both together we have m = p+2 and m=3p => 3p = p+2 => p = 1

So, m = p+2 = 3 And 1/p = 1 Clearly m is not equal to 1/p. So, lines are not parallel.

So, Answer will be C. Hope it helps!

You say that for two lines to be parallel their product should be 1. I don't get it, I thought that for two lines to be parallel they had to have the same slope. So in this case m = p

Please clarify Thanks Cheers J

For 2 lines to be parallel, their slopes must be equal and for 2 lines to be perpendicular, the product of slopes should = -1 Here the question asked whether the slope of the reflection of line y=px+q which isx=py+q parallel to line y= mx+n

So the slope of reflected line x=py+q ---------> 1/p

ie. is 1/p= m or in this case is pm= 1
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A line with the equation y = px + q is reflected over the [#permalink]

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21 Sep 2014, 19:52

jlgdr wrote:

nktdotgupta wrote:

A line with the equation y = px + q is reflected over the line y = x. Is the reflection of this line parallel to the line y = mx + n?

(1) m = p + 2

(2) m = 3p

When a line is reflected around y=x then the equation of the reflected line is got by swapping the x and y of the original equation. So, equation of the reflection of the line y=px + q will be x = py + q

So, for this line to be parallel to y = mx + n we just need to make sure that the slope of the two lines are equal i.e. 1/p = m

STAT1 m = p + 2 But we cannot say whether m=1/p will be true or not. Proof is below. Trying m =1/p we get 1/p = p + 2 p^2 + 2p - 1 =0 p = { -2 +- sqrt(4 + 4) }/2 = { -2 + 4sqrt2 } /2 But there can be many other values of p for which m = p+2 will be true but m = 1/p will not be true. So, INSUFFICIENT

STAT2 m = 3p But we cannot say whether m=1/p will be true or not. Proof is below. Trying m = 1/p 1/p = 3p => p^2 = 1/3 But there can be many other values of p for which m = p+2 will be true but m = 1/p will not be true. So, INSUFFICIENT

Taking both together we have m = p+2 and m=3p => 3p = p+2 => p = 1

So, m = p+2 = 3 And 1/p = 1 Clearly m is not equal to 1/p. So, lines are not parallel.

So, Answer will be C. Hope it helps!

You say that for two lines to be parallel their product should be 1. I don't get it, I thought that for two lines to be parallel they had to have the same slope. So in this case m = p

Please clarify Thanks Cheers J

Hi

Probably you are comparing with line equation y = mx+c, where slope is m, but this equation of the line is x= py+q, So if you change the equation to y = mx+c form, in the slope intercept form I mean, then we will get (x-q)/p = y y = x/p - q/p y = x(1/p) - q/p

hence the slope will be 1/p

Slope will change because we need to change the equation to make it in the form of slope intercept, if we want to express slope in terms of y =mx+c.. got it? let me know if there is still any confusion.. Cheers

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

A line with the equation y = px + q is reflected over the line y = x. Is the reflection of this line parallel to the line y = mx + n?

(1) m = p + 2

(2) m = 3p

Modify the original condition and the question. If y=px+q is reflected over the line y=x, x=py+q, py=x-q, y=(x/p)-(q/p), which is the question if it is parallel to the line y=mx+n. Therefore, 1/p=m? -> mp=1?. There are 4 variables(m,p,n,q), which should match with the number of equations. So, you need 4 equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer. In 1) & 2), p=1, m=3 -> mp=3=/1, which is no and sufficient. Therefore, the answer is C.

-> For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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