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- 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.

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- anonymous

- chestercat

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- anonymous

|dw:1327602494338:dw|

- anonymous

What do I do next? I mean how to find x and y?

- anonymous

we know that |dw:1327602954262:dw|

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- 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

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

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

sorry I made a mistake

- anonymous

I didn't get the last part. :O

- anonymous

|dw:1327604089388:dw|

- 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

|dw:1327604371124:dw|

- 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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