A line segment goes from #(3 ,1 )# to #(2 ,4 )#. The line segment is dilated about #(2 ,2 )# by a factor of #3#. Then the line segment is reflected across the lines #x = 4# and #y=-1#, in that order. How far are the new endpoints form the origin?

Question

A line segment goes from #(3  ,1  )# to #(2  ,4  )#.  The line segment is dilated about #(2  ,2  )# by a factor of #3#.  Then the line segment is reflected across the lines #x = 4# and #y=-1#, in that order.  How far are the new endpoints form the origin?

Sophia Alfred · Accepted Answer

#color(indigo)("Distances of A and B from origin after dilation and reflection "#

#color(indigo)(3.16, 12.65 " respectively"# #A (3,1), B (2,4), " dilated about C(2,2) by a factor of 3"# #A' (x,y) =3 * A(x,y) - C (x,y) =(( 9 ,3) - (2,2)) = (7,1)# #B'(x,y) = 3 ^ B (x,y) - C(x,y) = ((6,12) - (2,2)) = (4,10)# #color(brown)("reflect thru " x = 4, y = -1, h=4, k= -1. (2h-x, 2k-y)"# #A''(x,y) = (2h-x, 2k-y) = ((8 - 7), (-2-1)) = (1,-3)# #"similarly " B''(x,y) = ((8-4),(-2-10)) = (4, -12)# #bar(A''O) = sqrt(1^2 + 3^2) = sqrt 10 = 3.16# #bar(B''O) = sqrt(4^2 + 12^2) = sqrt 160 = 12.65#

Sadie Allen · Answer

undefined After dilating the line segment about the point (2, 2) by a factor of 3, the new endpoints are (9, 5) and (0, 11).

After reflecting across the line x = 4, the new endpoints become (5, 5) and (8, 11).

Then, reflecting across the line y = -1, the new endpoints are (5, 11) and (8, 5).

The distance of these endpoints from the origin can be calculated using the distance formula $ \sqrt{(x - x_0)^2 + (y - y_0)^2} $, where $ (x_0, y_0) $ is the origin (0, 0).

For the endpoint (5, 11):
$ \sqrt{(5 - 0)^2 + (11 - 0)^2} = \sqrt{25 + 121} = \sqrt{146} $

For the endpoint (8, 5):
$ \sqrt{(8 - 0)^2 + (5 - 0)^2} = \sqrt{64 + 25} = \sqrt{89} $

So, the distances from the new endpoints to the origin are approximately $ \sqrt{146} $ and $ \sqrt{89} $ units.