Answer to Question #97318 in Geometry for Max

Question #97318
Find the reflection of the point A(3, 1) in the line L with equation r · (2i − j) = 0.
Use your answer to find the reflection in L of the line r · (i − j) − 2 = 0.
1
Expert's answer
2019-10-28T11:30:14-0400

Reflection of a Point

We are going to fins the reflection of a point in the a Line

Solution:

Given, Point A (3, 1)


Let the Equation of r as


"\\vec r = x \\vec I + y \\vec j"

Equation of L is given as

"\\vec r.(2 \\vec i - \\vec j ) = 0\\\\\n (x \\vec i + y \\vec j). (2 \\vec i - \\vec j ) =0"

"2x - y = 0"

"(since \\space \n\\vec i . \\vec i = \\vec j . \\vec j = \\vec k .\\vec k = 1\\space and \\space \\space \n\\vec i . \\vec j = \\vec j . \\vec k = \\vec k . \\vec i =0)"

The reflection of point A(3, 1) lies on the  line perpendicular to the line "2x - y = 0"


Perpendicular line is

"x +2 y + k= 0"

The Point (3, 1) lies on this line

"3 +2( 1) + k = 0 \\\\\nk= -5"

Perpendicular line is

"x + 2y -5= 0"

The Intersecting Point of the Lines 2x−y=0 and x+2y−5=0 is the midpoint of the points (3, 1) and Reflecting point


Plug "y = 2x" in the equation "x + 2y - 5 = 0"


then

"x + 4x - 5 = 0\\\\\nx= 1 \\space and \\space y =2"

Let ( m, n ) is the reflection point

"so,"

"\\frac {3+m} {2} = 1 \\space and \\space \\frac {1+n} {2} = 2 \\\\\n m = -1 \\space and \\space n = 3"

Reflection Point is ( -1, 3)


(b).

Let

"L_1 : \\vec r .(\\vec i - \\vec j ) - 2=0\\\\\n (x \\vec i + y \\vec j) .(\\vec i - \\vec j ) - 2 = 0\\\\\n x - y - 2 = 0"

The intersecting point of these two line is (- 2, - 4)


Consider a point on the Line L1 as (2, 0)


Now, Let the image of the line L1 about the line L is L2


So, the image of the point (2, 0) will be lie on the line L2


Let the image of the point (2, 0) is (p, q)


So, "(\\frac { p+2} {2}, \\frac {q }{2} )" lies on the line L


"2 (\\frac { p+2} {2} ) - (\\frac {q }{2}) = 0"


"2p + 4 = q"

Slope of (p, q) and (2, 0) is

"\\frac {q} {p - 2} = \\frac {-1} {2}"

"p + 2q =2"

Solving the equations "2p + 4 = q \\space and \\space p + 2q = 2"

We get,

"p = \\frac {-6}{5} \\space and \\space q = \\frac {8}{5}"

Reflection in L Passes through ( -2, -4) and ("\\frac {-6}{5}, \\frac {8}{5}" )


"y - y_1 = \\frac {y_2 - y_1}{x_2 - x_1} (x - x_1) \\\\\n\ny + 4 = \\frac {\\frac {8}{5} + 4} {\\frac {-6} {5} +2} (x +2)"

"y + 4 = 7 (x +2)"

"Answer: y = 7x +10"


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