A line with the equation y = px + q is reflected over the

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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@bagrettin pls verify my reasoning. Thanks for confirming the answer

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Hello, gmat1220

Your reasoning is right!

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.

Hello bagrettin
Your explanation rocks!!!!! Thanks so much.

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Hello, gmat1220!

Happy to help you) You are always welcome!

Great explanation bagrettin !
Thanks to gmat1220 also.

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.

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


I know this question has been discussed earlier..bt that discussion didnt help me out..thats y have made new topic....

@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?

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.

Hope it helps!

Thanks alot..!!

i got it now..!

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

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

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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