Question

in exercises 11-14, use the translation. (x,y)→(x - 8,y + 4)

1. what is the image of a(2,6)?
2. what is the image of b(-1,5)?
3. what is the preimage of c(-3,-10)?
4. what is the preimage of d(4,-3)?

Explanation:

Step1: Find the image of a point using the given translation rule

For a point $(x,y)$ with translation $(x,y)\to(x - 8,y + 4)$.

Step2: Solve for the image of point A

Given $A(2,6)$, substitute $x = 2$ and $y=6$ into the translation rule.
$x'=2 - 8=-6$ and $y'=6 + 4 = 10$. So the image of $A(2,6)$ is $A'(-6,10)$.

Step3: Solve for the image of point B

Given $B(-1,5)$, substitute $x=-1$ and $y = 5$ into the translation rule.
$x'=-1-8=-9$ and $y'=5 + 4=9$. So the image of $B(-1,5)$ is $B'(-9,9)$.

Step4: Find the pre - image of a point

Let the pre - image of a point $(x',y')$ be $(x,y)$. We have the equations $x-8=x'$ and $y + 4=y'$. So $x=x'+8$ and $y=y'-4$.

Step5: Solve for the pre - image of point C'

Given $C'(-3,-10)$, substitute $x'=-3$ and $y'=-10$ into the pre - image equations.
$x=-3 + 8=5$ and $y=-10-4=-14$. So the pre - image of $C'(-3,-10)$ is $C(5,-14)$.

Step6: Solve for the pre - image of point D'

Given $D'(4,-3)$, substitute $x'=4$ and $y'=-3$ into the pre - image equations.
$x=4 + 8=12$ and $y=-3-4=-7$. So the pre - image of $D'(4,-3)$ is $D(12,-7)$.

Answer:

1. $A'(-6,10)$
2. $B'(-9,9)$
3. $C(5,-14)$
4. $D(12,-7)$