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EXAMPLE 3

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a. The vector is BC . From initial point B to terminal point C , you move 9 units right and 2 units down. So, the component form is 9, –2 . EXAMPLE 3. Identify vector components. Name the vector and write its component form. SOLUTION. b.

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Presentation Transcript
slide1
a.

The vector is BC . From initial point B to terminal point C, you move 9 units right and 2 units down. So, the component form is 9, –2 .

EXAMPLE 3

Identify vector components

Name the vector and write its component form.

SOLUTION

slide2
b.

The vector is ST . From initial point S to terminal point T, you move 8 units left and 0 units vertically. The component form is –8, 0 .

EXAMPLE 3

Identify vector components

Name the vector and write its component form.

SOLUTION

slide3
First, graph ∆ABC. Use 5, –1 to move each vertex 5 units to the right and 1 unit down. Label the image vertices. Draw ∆ A′B′C′. Notice that the vectors drawn from preimage to image vertices are parallel.

EXAMPLE 4

Use a vector to translate a figure

The vertices of ∆ABC are A(0, 3), B(2, 4), and C(1, 0). Translate∆ABC using the vector 5, –1 .

SOLUTION

slide4
SOLUTION

The vector is RS . From initial point R to terminal point S, you move 5 units right and 0 units vertically. The component form is 5, 0 .

for Examples 3 and 4

GUIDED PRACTICE

Name the vector and write its component form.

4.

slide5
The vector is TX . From initial point T to terminal point S, you move 0 units horizontally and 3 units up. The component form is 0, 3 .

for Examples 3 and 4

GUIDED PRACTICE

5.

Name the vector and write its component form.

SOLUTION

slide6
The vector is BK . From initial point B to terminal point K, you move 5 units left and 2 units up. So, the component form is –5 , 2 .

for Examples 3 and 4

GUIDED PRACTICE

Name the vector and write its component form.

6.

SOLUTION

slide7
for Examples 3 and 4

GUIDED PRACTICE

The vertices of ∆LMN are L(2, 2), M(5, 3), and N(9, 1). Translate ∆LMN using the vector –2, 6 .

7.

SOLUTION

Find the translation of each vertex by subtracting 2 from its x-coordinate and adding 6 to its y-coordinate.

(x, y) → (x – 2, y + 6)

L(2, 2) → L′(0, 8)

M(5, 3) → M′(3, 9)

N(9, 1) → N′(7, 7)

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