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
(1) m = p + 2
(2) m = 3p
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Coordinate geometry
Source: Beat The GMAT — Data Sufficiency |
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anshumishra
- Legendary Member
- Posts: 543
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Line L1 : y =px +q , has slope : s1 = pDeepthi 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 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
Thanks
Anshu
(Every mistake is a lesson learned )
Anshu
(Every mistake is a lesson learned )
great problem, an even better solution!! thanx!!
btw what is the source??
btw what is the source??
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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?
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?
-
anshumishra
- Legendary Member
- Posts: 543
- Thanked: 147 times
Your approach looks good to me.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?
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
Thanks
Anshu
(Every mistake is a lesson learned )
Anshu
(Every mistake is a lesson learned )
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)
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)
Last edited by clock60 on Sat Jan 15, 2011 1:10 pm, edited 1 time in total.
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anshumishra
- Legendary Member
- Posts: 543
- Thanked: 147 times
yeah, so for the two lines: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)
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.
Thanks
Anshu
(Every mistake is a lesson learned )
Anshu
(Every mistake is a lesson learned )
wow! that was superb!anshumishra wrote:Line L1 : y =px +q , has slope : s1 = pDeepthi 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 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
