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Re: A line with the equation y = px + q is reflected over the li [#permalink]
02 Oct 2012, 18:13

3

This post received KUDOS

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.

Re: A line with the equation y = px + q is reflected over the li [#permalink]
02 Oct 2012, 20:40

1

This post received KUDOS

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]
02 Oct 2012, 19:33

@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]
28 Nov 2013, 08:22

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]
29 Nov 2013, 02:56

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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Re: A line with the equation y = px + q is reflected over the li
[#permalink]
29 Nov 2013, 02:56