Question

the vertices of $\triangle def$ are $d(2, 5)$, $e(6, 3)$, and $f(4, 0)$. translate $\triangle def$ using the given vector. graph $\triangle def$ and its image. (see example 2.)

1. $\langle 6, 0 \

angle$

Explanation:

Step1: Define translation rule

For vector $\langle 6, 0
angle$, add 6 to $x$-coordinate, keep $y$-coordinate same: $(x,y) \to (x+6, y)$

Step2: Find new vertex $D'$

Substitute $D(2,5)$: $2+6=8$, $y=5$
$D'(8,5)$

Step3: Find new vertex $E'$

Substitute $E(6,3)$: $6+6=12$, $y=3$
$E'(12,3)$

Step4: Find new vertex $F'$

Substitute $F(4,0)$: $4+6=10$, $y=0$
$F'(10,0)$

Step5: Plot original and image

Plot $D(2,5), E(6,3), F(4,0)$ for $\triangle DEF$; plot $D'(8,5), E'(12,3), F'(10,0)$ for $\triangle D'E'F'$, then connect vertices for each triangle.

Answer:

Vertices of the translated triangle $\triangle D'E'F'$ are $D'(8, 5)$, $E'(12, 3)$, $F'(10, 0)$.
To graph:

1. Plot original points $D(2,5)$, $E(6,3)$, $F(4,0)$ and connect them to form $\triangle DEF$.
2. Plot translated points $D'(8,5)$, $E'(12,3)$, $F'(10,0)$ and connect them to form $\triangle D'E'F'$.