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

Question

1987 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-29T14:24:01

The new end points are $(4,16)$ and $(7,-5)$
The length of the new line segment is $=21.2$

Let the end points be $A=(2,8)$ and $B=(3,1)$

and $C=(1,4)$

Let $A'$ and $B'$ bethe new end points

Then,

$vec(CA')=3vec(CA)$

$=3*<2-1,8-4> =3*<1,4> = <3,12>$

$A'=(3,12)+(1,4)=(4,16)$

Similarly,

$vec(CB')=3vec(CB)$

$=3*<3-1,1-4> =3*<2,-3> = <6,-9>$

$B'=(6,-9)+(1,4)=(7,-5)$

The length of the line segment is

$A'B'=sqrt((7-4)^2+(-5-16)^2)$

$=sqrt((3)^2+(-21)^2)$

$=sqrt450$

$=21.2$

The length of the olb line segment is

$AB=sqrt((3-2)^2+(1-8)^2)$

$=sqrt(1+49)$

$=sqrt(50)$

$=7.07$

$A'B'=3*AB$

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