A triangle has corners at $(2, 7 )$ , $( 8, 3 )$ , and $( 4 , 8 )$ . If the triangle is dilated by $7 x$ around $(1, 3)$ , what will the new coordinates of its corners be?

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Answer 1 · 2017-03-14T09:20:31

The new coordinates are $(8,31)$ , $(50,0)$ and $(22,38)$

We can work this with vectors

Let the points be

$A=(2,7)$

$B=(8,3)$

$C=(4,8)$

$D=(1,3)$

$vec(DA)= <2-1,7-3> = <1,4>$

$vec(DB) = <8-1,3-3> = <7,0>$

$vec(DC)= <4-1, 8-3> = <3,5>$

Let $A'$ be the new point

Then

$vec(DA')=7vec(DA)=7+<1,4> = <7,28>$

So the coordinates of $A'=(1+7,3+28)=(8,31)$

Let $B'$ be another new point

Then,

$vec(DB')=7vec(DB)=7* <7,0> =<49,0>$

The coordinates of $B'=(1+49,3+0) =(50,0)$

Let $C'$ be another new point

Then,

$vec(DC')=7vec(DC)=7* <3,5> =<21,35>$

The coordinates of $C'=(1+21,3+35) =(22,38)$

A triangle has corners at (2, 7 )(2, 7 ), ( 8, 3 )( 8, 3 ), and ( 4 , 8 )( 4 , 8 ). If the triangle is dilated by 7 x 7 x around (1, 3)(1, 3), what will the new coordinates of its corners be?