A line segment has endpoints at (9 ,2 )(9,2) and (7 , 4)(7,4). The line segment is dilated by a factor of 3 3 around (1 , 5)(1,5). What are the new endpoints and length of the line segment?

1 Answer
Dean R.
May 13, 2018

(9,2) to (25, -4)(9,2)→(25,−4)

(7,4) to (19,2)(7,4)→(19,2)

New length 6 sqrt{2}6√2

Explanation:

There is indeed a Bolivia, United States, in North Carolina, not too far from Myrtle Beach.

I did the general case of this question [here].(https://socratic.org/questions/a-line-segment-has-endpoints-at-2-4-and-5-3-the-line-segment-is-dilated-by-a-fac-1)

I got for endpoints (a,b),(c,d),(a,b),(c,d), and factor rr around dilation point (p,q):(p,q):

(a,b) to ( (1-r)p + ra, (1-r)q+ rb) (a,b)→((1−r)p+ra,(1−r)q+rb), similarly for (c,d)(c,d),

new length l = r \sqrt{ (a-c)^2 + (b-d)^2 }l=r√(a−c)2+(b−d)2

These are old problems that probably no one ever looks at that I think are just here to give the noobs something to do. I'm an old timer at 26 days; I only answered because of Bolivia, United States.

I will now mindlessly substitute.

a=9,b=2,c=7,d=4,p=1,q=5,r=3a=9,b=2,c=7,d=4,p=1,q=5,r=3

(9,2) to ( (1-3)1 + 3(9), (1-3)5+ 3(2))= (25, -4)(9,2)→((1−3)1+3(9),(1−3)5+3(2))=(25,−4)

(7,4) to ( (1-3)1 + 3(7), (1-3)5+ 3(4)) = (19,2)(7,4)→((1−3)1+3(7),(1−3)5+3(4))=(19,2)

new length l = 3 \sqrt{ (9-7)^2 + (2-4)^2 } = 3 sqrt{8} =6 sqrt{2}l=3√(9−7)2+(2−4)2=3√8=6√2

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