Question

5.) graph $\triangle xyz$ with vertices $x(2, 4), y(6, 0),$ and $z(7, 2)$ and its image after the composition. (see exam21) translation: $(x,y) \to (x + 12,y + 4)$translation: $(x,y) \to (x - 5,y - 9)$describe the composition of translations.23)

Explanation:

For problem 21:

Step1: Apply first translation to X

$X'(2+12, 4+4) = (14, 8)$

Step2: Apply first translation to Y

$Y'(6+12, 0+4) = (18, 4)$

Step3: Apply first translation to Z

$Z'(7+12, 2+4) = (19, 6)$

Step4: Apply second translation to X'

$X''(14-5, 8-9) = (9, -1)$

Step5: Apply second translation to Y'

$Y''(18-5, 4-9) = (13, -5)$

Step6: Apply second translation to Z'

$Z''(19-5, 6-9) = (14, -3)$

Step7: Find net translation

Net $x$-shift: $12-5=7$, Net $y$-shift: $4-9=-5$

1. Identify the coordinates of the original triangle: $A(-4,2)$, $B(0,1)$, $C(-4,1)$
2. Identify coordinates after first translation: $A'(1,3)$, $B'(5,2)$, $C'(1,2)$: this is a shift right 5 units, up 1 unit, or $(x,y)\to(x+5,y+1)$
3. Identify coordinates after second translation: $A''(-3,-1)$, $B''(1,-2)$, $C''(-3,-2)$: this is a shift left 4 units, down 4 units from $A',B',C'$, or $(x,y)\to(x-4,y-4)$
4. Calculate the net translation: $(x+5-4,y+1-4)=(x+1,y-3)$

Answer:

• Pre-image vertices: $X(2,4)$, $Y(6,0)$, $Z(7,2)$
• Image after first translation: $X'(14,8)$, $Y'(18,4)$, $Z'(19,6)$
• Final image after second translation: $X''(9,-1)$, $Y''(13,-5)$, $Z''(14,-3)$
• The composition is equivalent to a single translation: $(x,y)\to(x+7,y-5)$

---

For problem 23: