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
d(_, _)
e(_, _)
f(_, _)
g(_, _)

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
l(_, _)
m(_, _)
n(_, _)

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

a) dilation with scale factor of 1/3 centered at the origin
b) translation along the vector ⟨-5, -1⟩
a(_, _)
b(_, _)
c(_, _)
d(_, _)
© gina wilson (all things algebra®, llc), 2015-20

Explanation:

Step 1: Solve part 1a (Translation of Rectangle DEFG)

For a translation rule \((x, y) \to (x + 6, y - 8)\), we apply this to each vertex:

• For \(D(-2, 7)\): \(x=-2 + 6 = 4\), \(y = 7-8=-1\), so \(D'(4, -1)\)
• For \(E(2, 3)\): \(x = 2+6 = 8\), \(y=3 - 8=-5\), so \(E'(8, -5)\)
• For \(F(0, 1)\): \(x = 0+6 = 6\), \(y=1 - 8=-7\), so \(F'(6, -7)\)
• For \(G(-4, 5)\): \(x=-4 + 6 = 2\), \(y=5 - 8=-3\), so \(G'(2, -3)\)

Step 1b: Reflection of translated rectangle in y - axis

The rule for reflection in the \(y\) - axis is \((x, y)\to(-x, y)\)

• For \(D'(4, -1)\): \(x=-4\), \(y = -1\), so \(D''(-4, -1)\)
• For \(E'(8, -5)\): \(x=-8\), \(y=-5\), so \(E''(-8, -5)\)
• For \(F'(6, -7)\): \(x=-6\), \(y=-7\), so \(F''(-6, -7)\)
• For \(G'(2, -3)\): \(x=-2\), \(y=-3\), so \(G''(-2, -3)\)

Step 2: Solve part 2a (Reflection of Triangle LMN in \(x = 5\))

The formula for reflection of a point \((x,y)\) in the line \(x = a\) is \((2a - x,y)\). Here \(a = 5\)

• For \(L(6,6)\): \(x=2\times5 - 6=4\), \(y = 6\), so \(L'(4,6)\)
• For \(M(8,8)\): \(x=2\times5 - 8 = 2\), \(y = 8\), so \(M'(2,8)\)
• For \(N(8,3)\): \(x=2\times5 - 8 = 2\), \(y = 3\), so \(N'(2,3)\)

Step 2b: \(270^{\circ}\) counter - clockwise rotation about the origin

The rule for \(270^{\circ}\) counter - clockwise rotation about the origin is \((x,y)\to(y,-x)\)

• For \(L'(4,6)\): \(x = 6\), \(y=-4\), so \(L''(6, - 4)\)
• For \(M'(2,8)\): \(x = 8\), \(y=-2\), so \(M''(8, - 2)\)
• For \(N'(2,3)\): \(x = 3\), \(y=-2\), so \(N''(3, - 2)\)

Step 3: Solve part 3a (Dilation of Quadrilateral ABCD with scale factor \(\frac{1}{3}\) centered at the origin)

The rule for dilation with scale factor \(k\) centered at the origin is \((x,y)\to(kx,ky)\). Here \(k=\frac{1}{3}\)

• For \(A(0,6)\): \(x = 0\times\frac{1}{3}=0\), \(y=6\times\frac{1}{3}=2\), so \(A'(0,2)\)
• For \(B(-3,-6)\): \(x=-3\times\frac{1}{3}=-1\), \(y=-6\times\frac{1}{3}=-2\), so \(B'(-1,-2)\)
• For \(C(-9,-6)\): \(x=-9\times\frac{1}{3}=-3\), \(y=-6\times\frac{1}{3}=-2\), so \(C'(-3,-2)\)
• For \(D(-12,-3)\): \(x=-12\times\frac{1}{3}=-4\), \(y=-3\times\frac{1}{3}=-1\), so \(D'(-4,-1)\)

Step 3b: Translation of dilated quadrilateral along vector \(\langle-5,-1

angle\)
The rule for translation along vector \(\langle h,k
angle\) is \((x,y)\to(x + h,y + k)\). Here \(h=-5\), \(k=-1\)

• For \(A'(0,2)\): \(x=0-5=-5\), \(y=2 - 1=1\), so \(A''(-5,1)\)
• For \(B'(-1,-2)\): \(x=-1-5=-6\), \(y=-2 - 1=-3\), so \(B''(-6,-3)\)
• For \(C'(-3,-2)\): \(x=-3-5=-8\), \(y=-2 - 1=-3\), so \(C''(-8,-3)\)
• For \(D'(-4,-1)\): \(x=-4-5=-9\), \(y=-1 - 1=-2\), so \(D''(-9,-2)\)

Answer:

1. a) After translation:

\(D'(4, - 1)\), \(E'(8, - 5)\), \(F'(6, - 7)\), \(G'(2, - 3)\)

1. b) After reflection:

\(D''(-4, - 1)\), \(E''(-8, - 5)\), \(F''(-6, - 7)\), \(G''(-2, - 3)\)

2. a) After reflection:

\(L'(4,6)\), \(M'(2,8)\), \(N'(2,3)\)

2. b) After rotation:

\(L''(6, - 4)\), \(M''(8, - 2)\), \(N''(3, - 2)\)

3. a) After dilation:

\(A'(0,2)\), \(B'(-1,-2)\), \(C'(-3,-2)\), \(D'(-4,-1)\)

3. b) After translation:

\(A''(-5,1)\), \(B''(-6,-3)\), \(C''(-8,-3)\), \(D''(-9,-2)\)