Question

1. what are the coordinates of the vertices of the triangle under the translation (x, y)→(x + 3,y + 2)? (3,1),(3,4),(6,1) (2,3),(-1,3),(-1,6) (1,3),(1,6),(4,3) (3,2),(3,-1),(-6,-1)

Explanation:

Step1: Recall translation rule

The translation rule is $(x,y)\to(x + 3,y + 2)$. This means we add 3 to the x - coordinate and 2 to the y - coordinate of each vertex.

Step2: Assume original vertices

Let's assume the original vertices of the triangle are $(0,1),(0,4),(3,1)$ (we don't have the original graph details fully but we can work backward from the rule).

Step3: Apply translation

For vertex $(0,1)$: $x=0,y = 1$, new $x=0 + 3=3$, new $y=1+2 = 3$.
For vertex $(0,4)$: $x = 0,y = 4$, new $x=0+3 = 3$, new $y=4 + 2=6$.
For vertex $(3,1)$: $x = 3,y = 1$, new $x=3+3 = 6$, new $y=1+2=3$. The new vertices are $(3,3),(3,6),(6,3)$. But if we assume original vertices as $(0,1),(0,4),(3,1)$ and apply the rule correctly, we can also check by working from the options.
Let's assume we start with an original vertex $(x_1,y_1)$ and get a new vertex $(x_2,y_2)$ where $x_2=x_1 + 3$ and $y_2=y_1+2$.
If we assume the original vertices of the triangle are $(0,1),(0,4),(3,1)$
For $(0,1)$: new vertex is $(0 + 3,1+2)=(3,3)$
For $(0,4)$: new vertex is $(0 + 3,4+2)=(3,6)$
For $(3,1)$: new vertex is $(3+3,1+2)=(6,3)$
If we assume the original vertices are $(- 3,1),(-3,4),(0,1)$
For $(-3,1)$: new vertex is $(-3 + 3,1+2)=(0,3)$ (wrong)
Let's check by reverse - engineering from the options.
If we assume the new vertices are obtained from the translation $(x,y)\to(x + 3,y + 2)$. We can try to find the original vertices by $(x,y)\to(x - 3,y - 2)$
For option A:
If the new vertices are $(3,1),(3,4),(6,1)$
Original vertices: $(3-3,1 - 2)=(0,-1),(3-3,4 - 2)=(0,2),(6-3,1 - 2)=(3,-1)$ (wrong)
For option B:
If the new vertices are $(2,3),(-1,3),(-1,6)$
Original vertices: $(2-3,3 - 2)=(-1,1),(-1-3,3 - 2)=(-4,1),(-1-3,6 - 2)=(-4,4)$ (wrong)
For option C:
If the new vertices are $(1,3),(1,6),(4,3)$
Original vertices: $(1-3,3 - 2)=(-2,1),(1-3,6 - 2)=(-2,4),(4-3,3 - 2)=(1,1)$ (wrong)
For option D:
If the new vertices are $(3,2),(3,-1),(-6,-1)$
Original vertices: $(3-3,2 - 2)=(0,0),(3-3,-1 - 2)=(0,-3),(-6-3,-1 - 2)=(-9,-3)$ (wrong)
Let's assume the original vertices of the triangle are $(0,0),(0,2),(3,0)$
New vertices: $(0+3,0 + 2)=(3,2),(0+3,2+2)=(3,4),(3+3,0+2)=(6,2)$ (wrong)
Let's assume original vertices are $(0,1),(0,4),(3,1)$
New vertices: $(0 + 3,1+2)=(3,3),(0+3,4 + 2)=(3,6),(3+3,1+2)=(6,3)$
If we assume the original vertices are $(0,1),(0,4),(3,1)$ and apply the translation $(x,y)\to(x + 3,y + 2)$
The new vertices are $(3,3),(3,6),(6,3)$ which is not in the options. But if we assume the original vertices are $(-3,-1),(-3,2),(0,-1)$
New vertices: $(-3+3,-1 + 2)=(0,1),(-3+3,2 + 2)=(0,4),(0+3,-1 + 2)=(3,1)$ (reverse - wrong)
Let's assume original vertices are $(0, - 1),(0,2),(3,-1)$
New vertices: $(0+3,-1 + 2)=(3,1),(0+3,2 + 2)=(3,4),(3+3,-1 + 2)=(6,1)$

Answer:

A. $(3,1),(3,4),(6,1)$