experts?

Deepthi Subbu 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

Line L1 : y =px +q , has slope : s1 = p
Line y = x has slope = 1
The perpendicular line to y=x will have a slope equal to : = -1
Given that, Line L2 is parallel to y = mx+n, so slope = s2 = m

Now the reflection of the line (y = px+q) will form the same angle with this perpendicular line =>
Hence,
tan (x) = tan (y) {Formula is : tan x = (m1-m2)/(1+m1*m2)}
=> (p+1)/(1-p) = -(1+m)/(1-m)
=> 2pm = 2
=> p = 1/m ? OR pm = 1 ?

Statement 1:
m = p+2 => mp = P^2+2p (can be or can't be equal to 1, as the roots of this quadratic is not imaginary, depending on the value of p)
Not sufficient

Statement 2:
m = 3p => mp = 3p^2 (This can be equal to 1 or not depending on the value of p)
Not sufficient

Combining 1 and 2 :
3p = p+2 => p = 1
So, m = 3p = 3
=> mp = 3*1 ≠ 1 -- Sufficient

Hence, C

great problem, an even better solution!! thanx!!

btw what is the source??

hi guys i have few problems about this not trivial (to me) problem
i think any reflection of the line y=px+q will looks like y=(-p)x+b. the main here is that angle of falling will equal to the angle of reflection with opposite sign
and if reflected line y=(-p)x+b will || to some other line in our case y = mx + n, the question is -p=x. or x+p=0?
(1) m=p+2, the values of m,p can be 1-(-1)=2 and -1+1=0 yes,
or p=1, m=3, 1+3=4 not equal to 0
insufficient
(2) m=3p is valid for m=p=0 yes, p=1, m=3 no
both 3p=p+2, p=1, m=3 1+3=4 not equal to 0
so suff
as for my questions, is my approach valid, for what we need x=y ( to me it does not matter what is line of reflection)
and what is oa?

clock60 wrote:hi guys i have few problems about this not trivial (to me) problem
i think any reflection of the line y=px+q will looks like y=(-p)x+b. the main here is that angle of falling will equal to the angle of reflection with opposite sign
and if reflected line y=(-p)x+b will || to some other line in our case y = mx + n, the question is -p=x. or x+p=0?
(1) m=p+2, the values of m,p can be 1-(-1)=2 and -1+1=0 yes,
or p=1, m=3, 1+3=4 not equal to 0
insufficient
(2) m=3p is valid for m=p=0 yes, p=1, m=3 no
both 3p=p+2, p=1, m=3 1+3=4 not equal to 0
so suff
as for my questions, is my approach valid, for what we need x=y ( to me it does not matter what is line of reflection)
and what is oa?

Your approach looks good to me.
The difference, we have in our solution is based on the understanding of "reflection". As shown in the diagram attached in my previous post, I consider one of the line as the light beam and the line (y=x) as a mirror, after that used some basic physics principle: https://www.physicsclassroom.com/mmedia/optics/lr.cfm

hi anshumishra
thank you for kind words i got you point
i have one small doubt but it does not refer directly to the problem,
in my solution i tried to estimate tangent of angles of lines y=(-p)x+b and y = mx + n, it happens that -p=/=m. so they can`t be ||
but what if b=n, and -p=m i mean the lines coinside with this the answer will be E, or i am digging to deep?
(i remember similar trap in one question)

clock60 wrote:hi anshumishra
thank you for kind words i got you point
i have one small doubt but it does not refer directly to the problem,
in my solution i tried to estimate tangent of angles of lines y=(-p)x+b and y = mx + n, it happens that -p=/=m. so they can be ||
but what if b=x, and -p=m i mean the lines coinside with this the answer will be E, or i am digging to deep?
(i remember similar trap in one question)

yeah, so for the two lines:
L1 : y = -px+b
L2 : y = mx+n
If , m = -p , then they can be parallel OR the same line (if in addition to m=-p, b=n).
You are right, this could be used to trap.

anshumishra wrote:

Deepthi Subbu 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

Line L1 : y =px +q , has slope : s1 = p
Line y = x has slope = 1
The perpendicular line to y=x will have a slope equal to : = -1
Given that, Line L2 is parallel to y = mx+n, so slope = s2 = m

Now the reflection of the line (y = px+q) will form the same angle with this perpendicular line =>
Hence,
tan (x) = tan (y) {Formula is : tan x = (m1-m2)/(1+m1*m2)}
=> (p+1)/(1-p) = -(1+m)/(1-m)
=> 2pm = 2
=> p = 1/m ? OR pm = 1 ?

Statement 1:
m = p+2 => mp = P^2+2p (can be or can't be equal to 1, as the roots of this quadratic is not imaginary, depending on the value of p)
Not sufficient

Statement 2:
m = 3p => mp = 3p^2 (This can be equal to 1 or not depending on the value of p)
Not sufficient

Combining 1 and 2 :
3p = p+2 => p = 1
So, m = 3p = 3
=> mp = 3*1 ≠ 1 -- Sufficient

Hence, C

wow! that was superb!