anonymous
  • anonymous
Can someone explain in steps how to solve this? Find the image of the point P(2,1) under a reflection in the line x+y=4.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1327602494338:dw|
anonymous
  • anonymous
What do I do next? I mean how to find x and y?
anonymous
  • anonymous
we know that |dw:1327602954262:dw|

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anonymous
  • anonymous
lines are perpendicular so their multiple of slope is -1 first lines' slope is -1 so second one's slope must be 1 m=(y-y1)/(x-x1) m=1 y1=1 x1=2 then y=x+1
anonymous
  • anonymous
yeas, the answer i think might be (3,2). Plug x=2 into x+y=4 t get y=2, and then plug y=1 into x+y = 4 to get x=3. Now you can show that the gradient of the line passing through (3,2) and (2,1) is perpendicular to the gradient of your starting line x+y=4 by noting that \[x+y=4 \Longrightarrow y=-x+4\] so the gradient is -1. On the other hand, the gradient of the line passing through (3,2) and (2,1) is given by \[\frac{2-1}{3-2} = \frac{1}{1} = 1\] The products of the gradients is 1x(-1) = -1
anonymous
  • anonymous
y=4-x y=x+1 so 4-x=x+1 => x=3/2 =>y=5/2 (3/2,5/2) is coordinate of intersection so x changes 2 to 1,5 then it will decrease 1,5 to 1 y changes 1to 2,5 then it will increase 2,5 to 4 so P(1,4)
anonymous
  • anonymous
sorry I made a mistake
anonymous
  • anonymous
I didn't get the last part. :O
anonymous
  • anonymous
|dw:1327604089388:dw|
anonymous
  • anonymous
y=4-x y=x-1 so => y=3/2 =>x=5/2 (5/2,3/2) is coordinate of intersection so x changes 2 to 2,5 then it will decrease 2,5 to 3 y changes 1to 1,5 then it will increase 1,5 to 2 so P(3,2)
anonymous
  • anonymous
|dw:1327604371124:dw|
anonymous
  • anonymous
The coordinates of intersection are: (5/2,3/2) so the distance from (2,1) to the mid point is equal to the distance from (3,2) to the midpoint: \[\sqrt{(5/2-2)^2 + (3/2-1)^2} = \sqrt{1} = \sqrt{(3-5/2)^2 + (2-3/2)^2}\] So the respective distances are equal, and both points: (2,1) and (3,2) lie on the line perpendicular to the mirror line, therefore they must be reflections of each other.

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