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

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Answer 1 · 2017-04-03T06:33:34

The new segments are $(5/2,3)$ and $(7/2,11/2)$
The new length of the segment is $=sqrt29/2$

Let the segment be $A=(1,4)$ and $B=(3,9)$

After dilatation, the segment becomes $(A'=(x_A,y_A))$ and $B'=(x_B,y_B)$

Let the center of dilatation be $D=(4,2)$

Factor of dilatation is $=1/2$

We do this calcutation with vectors

$vec(DA')=1/2vec(DA)$

$((x_A-4),(y_A-2))=1/2((1-4),(4-2))$

$((x_A),(y_A))=((4-3/2),(2+1))=((5/2),(3))$

Similarly,

$vec(DB')=1/2vec(DB)$

$((x_B-4),(y_B-2))=1/2((3-4),(9-2))$

$((x_B),(y_B))=((4-1/2),(2+7/2))=((7/2),(11/2))$

Original length of line segment is

$AB=sqrt((3-1)^2+(9-4)^2)$

$=sqrt(4+25)$

$=sqrt29$

New length of ine segment is

$A'B'=sqrt((7/2-5/2)^2+(11/2-3)^2)$

$=sqrt(1+25/4)$

$=sqrt29/2$

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