A line segment has endpoints at $(5, 5)$ $(5 ,5 )$ and $(3, 5)$ $(3 ,5 )$ . The line segment is dilated by a factor of $\frac{1}{2}$ $1/2$ around $(3, 2)$ $(3 , 2)$ . What are the new endpoints and length of the line segment?

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Answer 1 · 2017-06-28T10:14:14

The new endpoints are $(4, \frac{7}{2})$ $(4,7/2)$ and $(3, \frac{7}{2})$ $(3,7/2)$ and the length of the line segment is $= 1$ $=1$

Let the line segment be $A B$ $AB$

$A = (5, 5)$ $A=(5,5)$ and

$B = (3, 5)$ $B=(3,5)$

$C = (3, 2)$ $C=(3,2)$

The length of the segment $A B = 2$ $AB=2$

Let point $A'=(x,y)$ be the point after dilatation, then

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

$((x-3),(y-2))=1/2((5-3),(5-2))=1/2((2),(3))$

Therefore,

$x-3=1$ , $=>$ , $x=4$

$y-2=3/2$ , $=>$ , $y=7/2$

The point $A'=(4,7/2)$

Let point $B'=(p,q)$ be the point after dilatation, then

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

$((p-3),(q-2))=1/2((3-3),(5-2))=1/2((0),(3))$

Therefore,

$p-3=0$ , $=>$ , $p=3$

$q-2=3/2$ , $=>$ , $q=7/2$

The point $B'=(3,7/2)$

The length of the segment $A'B'=1$

A line segment has endpoints at (5 ,5 )(5,5)(5 ,5 ) and (3 ,5 )(3,5)(3 ,5 ). The line segment is dilated by a factor of 1/2 121/2 around (3 , 2)(3,2)(3 , 2). What are the new endpoints and length of the line segment?