A line segment goes from #(1 ,2 )# to #(4 ,7 )#. The line segment is reflected across #x=6#, reflected across #y=-1#, and then dilated about #(1 ,1 )# by a factor of #2#. How far are the new endpoints from the origin?
1 Answer
Original segment
is transformed into
The distances from the origin to the new endpoints are
Explanation:
-
Reflection of a point with coordinates
#(a_0,b_0)# relative to a line#x=6# (vertical line intersecting X-axis at coordinate#x=6# ) will be horizontally shifted into a new X-coordinate obtained by adding to an X-coordinate of the axis of symmetry (#x=6# ) the distance from it of the original X-coordinates (#6-a_0# ).
Y-coordinate remains the same in this transformation.
So, new coordinates are:
#(a_1,b_1) = (6+(6-a_0),b_0)=(12-a_0,b_0)# -
Reflection of a point with coordinates
#(a_1,b_1)# relative to a line#y=-1# (horizontal line intersecting Y-axis at coordinate#y=-1# ) will be vertically shifted into a new Y-coordinate obtained by adding to an Y-coordinate of the axis of symmetry (#y=-1# ) the distance from it of the original Y-coordinates (#-1-b_1# ).
X-coordinate remains the same in this transformation.
So, new coordinates are:
#(a_2,b_2) = (a_1,-1+(-1-b_1))=#
# = (a_1,-2-b_1)=(12-a_0,-2-b_0)# -
Dilation about a center point
#(1,1)# by a factor of#2# will transform a point#(a_2,b_2)# into
#(a_3,b_3) = (1+2(a_2-1),1+2(b_2-1)) =#
# = (1+2(12-a_0-1),1+2(-2-b_0-1)) =#
# = (23-2a_0, -5-2b_0)# -
Using this formula for both ends of our original segment
#AB# , where#A(1,2)# and#B=(4,7)# :
4.1.#(a_0=1, b_0=2)#
# rarr# #(a_3=23-2*1, b_3=-5-2*2) = #
# = (21, -9)#
4.2.#(a_0=4, b_0=7)#
# rarr# #(a_3=23-2*4, b_3=-5-2*7) = #
# = (15, -19)# -
The distance of each end of a new segment from the origin are
#d_A = sqrt((21)^2+(-9)^2) = sqrt(552) ~~22.8 #
#d_B = sqrt((15)^2+(-19)^2) = sqrt(586) ~~24.2 #
