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October 23, 2014

October 23, 2014

Posted by **Knights** on Monday, March 4, 2013 at 9:15pm.

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.

- Analytic Geometry - Reflecting points over lines -
**bobpursley**, Monday, March 4, 2013 at 9:21pmA 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

- Analytic Geometry - Reflecting points over lines -
**Reiny**, Monday, March 4, 2013 at 9:34pmAs 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)

- Analytic Geometry - Reflecting points over lines -
**Knights**, Monday, March 4, 2013 at 10:13pmTHANKS 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?

- Analytic Geometry - Reflecting points over lines -
**Reiny**, Monday, March 4, 2013 at 11:06pmI 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?

- Analytic Geometry - Reflecting points over lines -
**Knights**, Tuesday, March 5, 2013 at 9:04amThanks a lot, after following your suggestions I got (1,9).

- Analytic Geometry - Reflecting points over lines -
**Reiny**, Tuesday, March 5, 2013 at 10:24amno, 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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