A line segment has endpoints at #(2 ,7 )# and #(5 ,4 )#. The line segment is dilated by a factor of #3 # around #(4 ,3 )#. What are the new endpoints and length of the line segment?

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Given:

Scale Factor for dilation is #3#.

Useful observations involving Dilation:

Isometry refers to a linear transformation which preserves the length.

Dilation is NOT an isometry. It creates similar figures only.

Here #bar (AB)# is the pre-image and after dilation, #bar (A'B')# is called the image.

The absolute value of the scale factor (k),

with the constraint #0 < k < 1,#

reduces the line segment #bar (AB)#,

enlarges if otherwise.

Each point on the line segment #bar (AB)# will get 3 times as far from the Center of Dilation, #(4,3)# since the scale factor is #3#.

Dilation preserves the angle of measure.

Note that the pre-image and the image are parallel.

Observe that the points (center of dilation #color(red)C# , A and A') collinear.

And, the points (C, B and B') are also collinear.

#bar (AB) |\| bar (A'B')#, since we have congruent corresponding angles.

Also, from #C(4,3)#, move up 4 units on the y-axis, and 2 units left on the x-axis to reach the end-point #A(2,7)#.

Move (4 x 3 = 12 units) up on the y-axis, and (2 x 3 = 6 units) left on the x-axis tor reach the end-point of #A'B'(-2, 15)#.

Similarly,

from #C(4,3)#, move one unit up on the y-axis and one unit right on the x-axis, to reach point #B(5,4)#.

From #C(4,3)#, move (1 x 3 = 3 units) on the y-axis, (1 x 3 = 3 units) to the right on the x-axis, to reach the point #B'(7,6)#.

New end-points: #A'(-2, 15) and B'(7,6)#

Find the length of #bar (A'B')#, using distance formula:

#color(blue)(D = sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

#rArr D=sqrt[(7-(-2)^2)+(6-15)^2)#

#rArr D=sqrt(9^2+(-9)^2)#

#rArr D=sqrt(162)#

#rArr D~~ 12.72792#

#bar (A'B')~~"12.73 units"#

Hope this solution is helpful.