A line segment has endpoints at #(5 ,8 )# and #(2 , 1)#. The line segment is dilated by a factor of #3 # around #(6 , 4)#. What are the new endpoints and length of the line segment?

Question

A line segment has endpoints at #(5  ,8  )# and #(2  , 1)#.  The line segment is dilated by a factor of #3  # around #(6 , 4)#.  What are the new endpoints and length of the line segment?

Savannah Aguilar · Accepted Answer

#color(orange)("New coordinates are " (3,16), (-6, -5)#

#color(brown)("Length of the line segment after dilation " vec(A’B’) ~~ 22.8473# #A(5, 8), B(2, 1), C(6, 4)#, dilation factor #3# #A’ = 3a - 2c# #A’((x),(y)) = 3 * ((5),(8)) - 2 * ((6),(4))# #A’((x),(y)) = ((3),(16))# #B’((x),(y)) = 3 * ((2),(1)) - 2 * ((6),(4))# #B’((x),(y)) = ((-6),(-5))# #color(brown)("Length of the line segment after dilation "# #color(green)(vec(A’B’) = sqrt((3- -6)^2 + (16- -5)^2) ~~ 22.8473#

Allison Allen · Answer

undefined To find the new endpoints of the line segment after dilation by a factor of 3 around the point (6, 4), follow these steps:

1. Find the vector representing the original line segment by subtracting the coordinates of one endpoint from the coordinates of the other endpoint.
2. Multiply this vector by the dilation factor (3) to obtain the vector representing the dilated line segment.
3. Add the coordinates of the center of dilation (6, 4) to this dilated vector to find the coordinates of the new endpoints.

Let's go through these steps:

Original line segment:
Endpoint 1: (5, 8)
Endpoint 2: (2, 1)

Vector representing the original line segment:
v = (2 - 5, 1 - 8) = (-3, -7)

Dilated vector:
v_dilated = (3 * -3, 3 * -7) = (-9, -21)

New endpoints:
Endpoint 1: (6, 4) + (-9, -21) = (-3, -17)
Endpoint 2: (6, 4) + (-9, -21) = (-3, -17)

So, the new endpoints of the line segment after dilation are (-3, -17) and (-3, -17).

To find the length of the dilated line segment, you can use the distance formula:

Distance = √((x2 - x1)^2 + (y2 - y1)^2)

For the new endpoints (-3, -17) and (-3, -17):

Length = √((-3 - (-3))^2 + (-17 - (-17))^2)
       = √(0^2 + 0^2)
       = √(0)
       = 0

Therefore, the length of the dilated line segment is 0.