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
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
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 + 4 = 7 (x +2)"
"Answer: y = 7x +10"
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