Question

1. find the pre - image (x - 9,y + 13)

g______→g(5, - 29)
h______→h(20, - 19)
i______→i(21, - 4)

1. find the pre - image (x + 7,y - 19)

g______→g(2,18)
h______→h(13,29)
i______→i(24,37)

1. translate the image.

translation: (x,y)→(x - 3,y - 5)

1. translate the image.

translation: (x,y)→(x - 2,y - 4)

Explanation:

3.

Step1: Set up equations for x - coordinate

For the transformation $(x,y)\to(x - 9,y + 13)$, if the image - point has coordinates $(x',y')$, then $x'=x - 9$ and $y'=y + 13$. For $G'(5,-29)$:
$x-9 = 5$, so $x=5 + 9=14$.
$y + 13=-29$, so $y=-29 - 13=-42$. So $G(14,-42)$.

Step2: For $H'(20,-19)$

$x-9 = 20$, so $x=20 + 9 = 29$.
$y + 13=-19$, so $y=-19 - 13=-32$. So $H(29,-32)$.

Step3: For $I'(21,-4)$

$x-9 = 21$, so $x=21+9 = 30$.
$y + 13=-4$, so $y=-4 - 13=-17$. So $I(30,-17)$.

Step1: Set up equations for x - coordinate

For the transformation $(x,y)\to(x + 7,y - 19)$, if the image - point has coordinates $(x',y')$, then $x'=x + 7$ and $y'=y - 19$. For $G'(2,18)$:
$x + 7=2$, so $x=2 - 7=-5$.
$y-19 = 18$, so $y=18 + 19=37$. So $G(-5,37)$.

Step2: For $H'(13,29)$

$x + 7=13$, so $x=13 - 7 = 6$.
$y-19 = 29$, so $y=29 + 19=48$. So $H(6,48)$.

Step3: For $I'(24,37)$

$x + 7=24$, so $x=24 - 7 = 17$.
$y-19 = 37$, so $y=37 + 19=56$. So $I(17,56)$

Step1: Apply translation to $R$

For the translation $(x,y)\to(x - 3,y - 5)$, when $x = 0,y = 0$ (coordinates of $R$), the new coordinates $R'$ are:
$x_R'=0 - 3=-3$, $y_R'=0 - 5=-5$. So $R'(-3,-5)$.

Step2: Apply translation to $S$

When $x = 3,y = 0$ (coordinates of $S$), the new coordinates $S'$ are:
$x_S'=3 - 3=0$, $y_S'=0 - 5=-5$. So $S'(0,-5)$.

Step3: Apply translation to $Z$

When $x = 0,y = 3$ (coordinates of $Z$), the new coordinates $Z'$ are:
$x_Z'=0 - 3=-3$, $y_Z'=3 - 5=-2$. So $Z'(-3,-2)$.

Answer:

$G(14,-42)$
$H(29,-32)$
$I(30,-17)$

4.