Analytic Geometry  Reflecting points over lines
posted by Knights on .
Let P = (5,1), and let Q be the reflection of P over the line y = 1/2x + 2. Find the coordinates of Q.
I don't understand how to start? Should we draw perpendicular lines?
Analytic Geometry  Reflecting points over lines  Steve, Monday, March 4, 2013 at 12:15pm
Given (x,y) and a line y = ax + c we want the point (x', y') reflected on the line.
Set d:= (x + (y  c)*a)/(1 + a^2)
Then x' = 2*d  x
and y' = 2*d*a  y + 2c
This relies on the fact the the distance from (h,k) to the line ax+by+c = 0 is
ah+bk+c/√(a^2+b^2)
Question: How did you get this? I don't understand.

A neat trick to see what Steve said is true...the distance from the line is the same.
Do this. TAke a piece of graph paper, plot the line, and the point. Then fold the paper along the line of symettry you drew, punch with a pin through the marked point so it goes through both sides of the paper. Unfold it. Notice the original point, and the new hole exactly the same distance from the line.
That is the "reflecton" of the point. Now reread Steves' solution 
As Steve told you, it uses the formula for the distance from a point to a line.
Here is another approach, perhaps it uses simpler algebra
let the reflected point be Q(a,b)
slope of given line is 1/2
slope of PQ = (b5)/(a1)
but they are perpendicular, so
(b5)/(a1) = 2
which after crossmultiplying and simplifying gives me
b = 72a
Clearly the midpoint of PQ must lie on the given line
midpoint is ( (a+1)/2 , (b+5)/2)
= ( (a+1)/2, (122a)/2 )
subbing into the equation
y = (1/2)x + 2
(122a)/2 = (1/2)((a+1)/2 + 2
times 2
12  2a = (a+1)/2 + 4
times 2 again
24  4a = a+1 + 8
5a = 15
a = 3 , then b = 76 = 1
the point P is (3,1) 
THANKS A LOT EVERYONE but if the point Q is on 3,1 (I suppose that is what you meant) then i don't think its right since if it is reflected over the line 1/2x+2 then it would be on the other side, right?

I really have to get new glasses, lol
but, ...
why don't you just change the numbers from (1,5) to (1,3) and follow the same steps? 
Thanks a lot, after following your suggestions I got (1,9).

no, that is not correct ...
my error was that I read your P(5,1) as (1,5)
so let's go back and change my steps to :
let the reflected point be Q(a,b)
slope of given line is 1/2
slope of PQ = (b1)/(a5)
but they are perpendicular, so
(b1)/(a5) = 2
which after crossmultiplying and simplifying gives me
b = 112a
Clearly the midpoint of PQ must lie on the given line
midpoint is ( (a+5)/2 , (b+1)/2)
= ( (a+5)/2, (122a)/2 )
subbing into the equation
y = (1/2)x + 2
(122a)/2 = (1/2)((a+5)/2 + 2
times 2
12  2a = (a+5)/2 + 4
times 2 again
24  4a = a+5 + 8
5a = 11
a = 11/5 = 2.2 , then b = 114.4 = 6.6
the point P is (2.2, 6.6)
check:
slope PQ = (6.6  1)/(2.2  5) = 5.6/2.8 = 2 , which is correct
distance for P(5,1) to line is
5  2 + 4/√(1^2+2^2) = 7/√5
distance for Q(2.2 , 6.6) to the line is
2.2  13.2 + 4/√5 = 7/√5
The point P(5,1) is reflected in the line
y = (1/2)x + 2 to the point Q(2.2 , 6.6) or Q(11/5 , 33/5)