Question

directions: graph and label each figure and its image under the sequence of transformations. give the coordinates of the image.

1. rectangle defg with vertices d(-2,7), e(2,3), f(0,1), and g(-4,5):

a) translation along the rule (x,y)→(x + 6,y - 8)
b) reflection in the y - axis

1. triangle lmn with vertices l(6,6), m(8,8), and n(8,3):

a) reflection in the line x = 5
b) 270° counterclockwise rotation about the origin

1. quadrilateral abcd with vertices a(0,6), b(-3,-6), c(-9,-6), and d(-12,-3):

a) partially obscured transformation
b) translation along the vector ⟨-5,-1⟩

Explanation:

Step1: Perform translation for rectangle DEFG

For point $D(-2,7)$:
$x=-2,y = 7$, after translation $(x,y)\to(x + 6,y-8)$, we have $x'=-2 + 6=4,y'=7-8=-1$, so $D'=(4,-1)$.
For point $E(2,3)$:
$x = 2,y=3$, after translation $x'=2 + 6=8,y'=3-8=-5$, so $E'=(8,-5)$.
For point $F(0,1)$:
$x = 0,y = 1$, after translation $x'=0+6 = 6,y'=1 - 8=-7$, so $F'=(6,-7)$.
For point $G(-4,5)$:
$x=-4,y = 5$, after translation $x'=-4+6 = 2,y'=5-8=-3$, so $G'=(2,-3)$.

Step2: Reflect the translated points of rectangle DEFG in the y - axis

The rule for reflection in the y - axis is $(x,y)\to(-x,y)$.
For $D'(4,-1)$, the new point $D''=(-4,-1)$.
For $E'(8,-5)$, the new point $E''=(-8,-5)$.
For $F'(6,-7)$, the new point $F''=(-6,-7)$.
For $G'(2,-3)$, the new point $G''=(-2,-3)$.

Step3: Perform reflection for triangle LMN in the line $x = 5$

The distance between a point $(x,y)$ and the line $x = 5$ is $d=\vert x - 5\vert$.
For point $L(6,6)$:
The distance $d=\vert6 - 5\vert=1$. The new x - coordinate is $x'=5-(6 - 5)=4$, and $y'=6$, so $L'=(4,6)$.
For point $M(8,8)$:
The distance $d=\vert8 - 5\vert=3$. The new x - coordinate is $x'=5-(8 - 5)=2$, and $y'=8$, so $M'=(2,8)$.
For point $N(8,3)$:
The distance $d=\vert8 - 5\vert=3$. The new x - coordinate is $x'=5-(8 - 5)=2$, and $y'=3$, so $N'=(2,3)$.

Step4: Rotate the reflected points of triangle LMN 270° counter - clockwise about the origin

The rule for a 270° counter - clockwise rotation about the origin is $(x,y)\to(y,-x)$.
For $L'(4,6)$, the new point $L''=(6,-4)$.
For $M'(2,8)$, the new point $M''=(8,-2)$.
For $N'(2,3)$, the new point $N''=(3,-2)$.

Step5: Assume the first transformation for quadrilateral ABCD is a 90° clockwise rotation about the origin (since the first part is not clear, assuming a common transformation)

The rule for a 90° clockwise rotation about the origin is $(x,y)\to(y,-x)$.
For point $A(0,6)$: $A'=(6,0)$.
For point $B(-3,-6)$: $B'=(-6,3)$.
For point $C(-9,-6)$: $C'=(-6,9)$.
For point $D(-12,-3)$: $D'=(-3,12)$.

Step6: Translate the rotated points of quadrilateral ABCD along the vector $\langle-5,-1

angle$
The rule for translation along the vector $\langle a,b
angle$ is $(x,y)\to(x + a,y + b)$ with $a=-5,b=-1$.
For $A'(6,0)$: $x'=6-5 = 1,y'=0-1=-1$, so $A''=(1,-1)$.
For $B'(-6,3)$: $x'=-6-5=-11,y'=3-1 = 2$, so $B''=(-11,2)$.
For $C'(-6,9)$: $x'=-6-5=-11,y'=9-1 = 8$, so $C''=(-11,8)$.
For $D'(-3,12)$: $x'=-3-5=-8,y'=12-1 = 11$, so $D''=(-8,11)$.

Answer:

1. $D''(-4,-1),E''(-8,-5),F''(-6,-7),G''(-2,-3)$
2. $L''(6,-4),M''(8,-2),N''(3,-2)$
3. $A''(1,-1),B''(-11,2),C''(-11,8),D''(-8,11)$