Posted by Knights on Monday, March 4, 2013 at 9:15pm.
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 thru the marked point so it goes thru 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
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 thru the marked point so it goes thru 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 = (b-5)/(a-1)
but they are perpendicular, so
(b-5)/(a-1) = -2
which after cross-multiplying and simplifying gives me
b = 7-2a
Clearly the midpoint of PQ must lie on the given line
midpoint is ( (a+1)/2 , (b+5)/2)
= ( (a+1)/2, (12-2a)/2 )
subbing into the equation
y = (1/2)x + 2
(12-2a)/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 = 7-6 = 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 dont 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 = (b-1)/(a-5)
but they are perpendicular, so
(b-1)/(a-5) = -2
which after cross-multiplying and simplifying gives me
b = 11-2a
Clearly the midpoint of PQ must lie on the given line
midpoint is ( (a+5)/2 , (b+1)/2)
= ( (a+5)/2, (12-2a)/2 )
subbing into the equation
y = (1/2)x + 2
(12-2a)/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 = 11-4.4 = 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)
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