Analytic Geometry - Reflecting points over lines

posted by .

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.

  • Analytic Geometry - Reflecting points over lines -

    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

  • Analytic Geometry - Reflecting points over lines -

    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)

  • Analytic Geometry - Reflecting points over lines -

    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?

  • Analytic Geometry - Reflecting points over lines -

    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?

  • Analytic Geometry - Reflecting points over lines -

    Thanks a lot, after following your suggestions I got (1,9).

  • Analytic Geometry - Reflecting points over lines -

    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)

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. algebra

    Find the slope of the line through the given points. (2,4) (6,-3). m=y2-y1 over x2-x1. (-3)-(4) over(6)-(2) -3-4 over 6-2 -7 over 4. I plotted the two coordinates (2,-4) (6)(-3), and the slope is -7 over 4, right?
  2. Geometry

    Which shows the correct coordinates for the reflection of the point over the line y = -4 a. (-6,9)->(-6, 17) b. (-6,9)->(6,-17) c. (-6,9)->(-10,-9) d. (-6,9)->(2,-9) I how to reflect over a line. Can you help me please?
  3. Math

    After a reflection over the x-axis I have to find the coordinates of A1, B1, C1 I am having trouble with b1. A(1,4) B(3,-2) C(4,2) The reflection should be: A1 should be (1,-4) B1 ?
  4. Analytic Geometry - Reflecting points over lines

    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?
  5. Algebra/Geometry

    Identify the sequence of transformations that maps quadrilateral abcd onto quadrilateral a"b"c"d" Answers 180 rotation around the origin; reflection over the x-axis translation (x,y) -> (x - 2, y + 0); reflection over the line x …
  6. analytic geometry

    The image of a point after a reflection over the line y = –x is (7, –1). Find the coordinates of the preimage.
  7. Math

    Which transformation will be equivalent to rotating a figure 90° counterclockwise?
  8. Math

    Which transformation will be equivalent to rotating a figure 270° counterclockwise?
  9. math

    which pair of transformations to the figure shown below would produce an image that is on top of the original A. a translation to the right and a reflection over the vertical line of reflection shown. B. a translation down and a reflection …
  10. math

    which pair of transformations to the figure shown below would produce an image that is on top of the original A. a translation to the right and a reflection over the vertical line of reflection shown. B. a translation down and a reflection …

More Similar Questions