Question

1. △pqr has vertices p(0, -4),q(1, 3) and r(2, -5). find the vertices of p q r after a composition of transformations in the order they are listed.

translation: (x, y)→(x, y + 3)
translation: (x, y)→(x - 1, y - 1)
p(0,-4) p(__,) p(,__)
q(1,3) q(__,) q(,__)
r(2,-5) r(__,) r(,__)

Explanation:

Step1: Apply first translation to point P

For point P(0, - 4), using the translation $(x,y)\to(x,y + 3)$, we substitute $x = 0$ and $y=-4$. So $P'(0,-4 + 3)=P'(0,-1)$.

Step2: Apply second translation to P'

Using the translation $(x,y)\to(x - 1,y - 1)$ on $P'(0,-1)$, we substitute $x = 0$ and $y=-1$. Then $P''(0-1,-1 - 1)=P''(-1,-2)$.

Step3: Apply first translation to point Q

For point Q(1,3), using the translation $(x,y)\to(x,y + 3)$, we substitute $x = 1$ and $y = 3$. So $Q'(1,3 + 3)=Q'(1,6)$.

Step4: Apply second translation to Q'

Using the translation $(x,y)\to(x - 1,y - 1)$ on $Q'(1,6)$, we substitute $x = 1$ and $y = 6$. Then $Q''(1-1,6 - 1)=Q''(0,5)$.

Step5: Apply first translation to point R

For point R(2,-5), using the translation $(x,y)\to(x,y + 3)$, we substitute $x = 2$ and $y=-5$. So $R'(2,-5 + 3)=R'(2,-2)$.

Step6: Apply second translation to R'

Using the translation $(x,y)\to(x - 1,y - 1)$ on $R'(2,-2)$, we substitute $x = 2$ and $y=-2$. Then $R''(2-1,-2 - 1)=R''(1,-3)$.

Answer:

P'(0,-1) P''(-1,-2)
Q'(1,6) Q''(0,5)
R'(2,-2) R''(1,-3)