findth e im ag e o f th eli n e. y=3x+4.after.reflectioneintheliney=2x -3?
4 answers
Y=91\63x -434\63
This article will probably help confirm your answer
http://www.sdmath.com/math/geometry/reflection_across_line.html
http://www.sdmath.com/math/geometry/reflection_across_line.html
I plotted your 3 lines.
Looks like your slope is off, but your common point looks correct
www.wolframalpha.com/input/?i=plot+y%3D3x%2B4,+y%3D2x+-3+,+y%3D91%2F63x+-434%2F63
Looks like your slope is off, but your common point looks correct
www.wolframalpha.com/input/?i=plot+y%3D3x%2B4,+y%3D2x+-3+,+y%3D91%2F63x+-434%2F63
suppose you want to reflect the line L2:y=cx+d across the line L1:y=ax+b,
creating a new line L3: y=mx+n
You know they intersect at some point (h,k), in this case, (-7,-17)
Consider the slopes of L1 and L2. They form angles A and C with the x-axis, such that
tanA = a
tanC = c
Our new line L3 will form an angle M with the x-axis, so that tanM = m
Now, we want to find M such that C-A = A-M
That is, M = 2A-C
m = tanM = tan(2A-C)
= (tan2A -tanC)/(1 + tanC tan2A)
= (tan2A - c)/(1 + c*tan2A)
Now, tan2A = 2a/(1-a^2), so
m = ([2a/(1-a^2)] - c)/(1+c[2a/(1-a^2)])
m = -(a^2c + 2a - c)/(a^2 - 2ac - 1)
For our problem, a=2 and c=3, so
m = 13/9
Using the point-slope form of the line, our reflected line L3 has equation
y+17 = 13/9 (x+7)
creating a new line L3: y=mx+n
You know they intersect at some point (h,k), in this case, (-7,-17)
Consider the slopes of L1 and L2. They form angles A and C with the x-axis, such that
tanA = a
tanC = c
Our new line L3 will form an angle M with the x-axis, so that tanM = m
Now, we want to find M such that C-A = A-M
That is, M = 2A-C
m = tanM = tan(2A-C)
= (tan2A -tanC)/(1 + tanC tan2A)
= (tan2A - c)/(1 + c*tan2A)
Now, tan2A = 2a/(1-a^2), so
m = ([2a/(1-a^2)] - c)/(1+c[2a/(1-a^2)])
m = -(a^2c + 2a - c)/(a^2 - 2ac - 1)
For our problem, a=2 and c=3, so
m = 13/9
Using the point-slope form of the line, our reflected line L3 has equation
y+17 = 13/9 (x+7)