Question

1. the translation $t_{(4,-5)}(\triangle mnp)$ indicates a transformation that moves every point of the preimage, $\triangle mnp$, right 4 units, and units to form the image, $\triangle mnp$.

$t_{(4,-5)}(\triangle mnp)$

1. which is the image of $\triangle jkl$?

$t_{(-2,1)}(\triangle jkl)$

1. deshawn wrote the following rule for the translation shown. what was his error?

$t_{(-4,-3)}(\triangle jkl)$

1. give the coordinates of the image.

$t_{(3,-2)}(\triangle abc)$ for $a(4,1), b(-3,2), c(4,-5)$
$a(7,-1), b$ _, $c$ _
$t_{(-5,0)}(\triangle def)$ for $d(4,4), e(-3,5), f(0,7)$
$d(-1,4), e$ _, $f$ _
$t_{(-8,-5)}(\triangle ghj)$ for $g(0,0), h(4,3), j(9,7)$
$g(-8,-5), h$ _, $j$ _

Explanation:

Step1: Translate point \( B(-3,2) \)

For a translation \( T_{(3,-2)} \), we add \( 3 \) to the \( x \)-coordinate and subtract \( 2 \) from the \( y \)-coordinate.
\( x=-3 + 3=0 \), \( y = 2-2=0 \). So \( B'=(0,0) \).

Step2: Translate point \( C(4,-5) \)

Add \( 3 \) to the \( x \)-coordinate and subtract \( 2 \) from the \( y \)-coordinate.
\( x = 4+3=7 \), \( y=-5 - 2=-7 \). So \( C'=(7,-7) \).

Step3: Translate point \( E(-3,5) \) for \( T_{(-5,0)} \)

Add \( - 5 \) to the \( x \)-coordinate and \( 0 \) to the \( y \)-coordinate.
\( x=-3-5=-8 \), \( y = 5+0=5 \). So \( E'=(-8,5) \).

Step4: Translate point \( F(0,7) \) for \( T_{(-5,0)} \)

Add \( -5 \) to the \( x \)-coordinate and \( 0 \) to the \( y \)-coordinate.
\( x=0 - 5=-5 \), \( y=7+0=7 \). So \( F'=(-5,7) \).

Step5: Translate point \( H(4,3) \) for \( T_{(-8,-5)} \)

Add \( -8 \) to the \( x \)-coordinate and \( -5 \) to the \( y \)-coordinate.
\( x=4-8=-4 \), \( y=3-5=-2 \). So \( H'=(-4,-2) \).

Step6: Translate point \( J(9,7) \) for \( T_{(-8,-5)} \)

Add \( -8 \) to the \( x \)-coordinate and \( -5 \) to the \( y \)-coordinate.
\( x=9-8 = 1 \), \( y=7-5=2 \). So \( J'=(1,2) \).

Answer:

\( B'=(0,0) \), \( C'=(7,-7) \), \( E'=(-8,5) \), \( F'=(-5,7) \), \( H'=(-4,-2) \), \( J'=(1,2) \)