Question

for exercises 11 and 12, trapezoid wxyz has vertices w(-1, 4), x(2, 3), y(2, 0), and z(-4, 2). write the coordinates of trapezoid wxyz after each translation. 11. 4 units left and 1 unit down 12. 3 units right and 6 units down

Explanation:

Step1: Recall translation rule

For a translation $a$ units left and $b$ units down, the rule for a point $(x,y)$ is $(x - a,y - b)$. For a translation $a$ units right and $b$ units down, the rule is $(x + a,y - b)$.

Step2: Calculate coordinates for 11

For point $W(-1,4)$: $x=-1,y = 4,a = 4,b = 1$. New - $x=-1-4=-5$, new - $y=4 - 1=3$, so $W'(-5,3)$.
For point $X(2,3)$: $x = 2,y = 3,a = 4,b = 1$. New - $x=2-4=-2$, new - $y=3 - 1=2$, so $X'(-2,2)$.
For point $Y(2,0)$: $x = 2,y = 0,a = 4,b = 1$. New - $x=2-4=-2$, new - $y=0 - 1=-1$, so $Y'(-2,-1)$.
For point $Z(-4,2)$: $x=-4,y = 2,a = 4,b = 1$. New - $x=-4-4=-8$, new - $y=2 - 1=1$, so $Z'(-8,1)$.

Step3: Calculate coordinates for 12

For point $W(-1,4)$: $x=-1,y = 4,a = 3,b = 6$. New - $x=-1 + 3=2$, new - $y=4-6=-2$, so $W'(2,-2)$.
For point $X(2,3)$: $x = 2,y = 3,a = 3,b = 6$. New - $x=2 + 3=5$, new - $y=3-6=-3$, so $X'(5,-3)$.
For point $Y(2,0)$: $x = 2,y = 0,a = 3,b = 6$. New - $x=2 + 3=5$, new - $y=0-6=-6$, so $Y'(5,-6)$.
For point $Z(-4,2)$: $x=-4,y = 2,a = 3,b = 6$. New - $x=-4 + 3=-1$, new - $y=2-6=-4$, so $Z'(-1,-4)$.

Answer:

1. $W'(-5,3),X'(-2,2),Y'(-2,-1),Z'(-8,1)$
2. $W'(2,-2),X'(5,-3),Y'(5,-6),Z'(-1,-4)$