Question

9.5 compositions of transformations
homework
name
date
per
for each point, give the coordinates of its image under the translation.

1. given points ( a(1,2) ), ( b(5,4) ), and ( c(3,0) ), graph the image of ( \triangle abc ) under the composition of transformations in the order listed.

translation: ( (x,y)\to(x - 4,y) )
reflection: in the ( x )-axis
( a(__) ), ( b() ), ( c(__) )
( a(__) ), ( b() ), ( c(__) )

1. given points ( a(1,2) ), ( b(5,4) ), and ( c(3,0) ), graph the image of ( \triangle abc ) under the composition of transformations in the order listed.

translation: ( langle 0,-3
angle )
reflection: in the ( y )-axis
( a(__) ), ( b() ), ( c(__) )
( a(__) ), ( b() ), ( c(__) )

1. given points ( a(-3,-5) ), ( b(0,-5) ), and ( c(1,-2) ), graph the image of ( \triangle abc ) under the composition of transformations in the order listed.

translation: ( langle 4,5
angle )
reflection: in the line ( y = x )
( a(__) ), ( b() ), ( c(__) )
( a(__) ), ( b() ), ( c(__) )

1. given points ( a(-1,-3) ), ( b(-4,-2) ), and ( c(0,2) ), graph the image of ( \triangle abc ) under the composition of transformations in the order listed.

reflection: in the line ( x = 1 )
translation: ( (x,y)\to(x - 2,y + 3) )
( a(__) ), ( b() ), ( c(__) )
( a(__) ), ( b() ), ( c(__) )

Explanation:

Problem 1

Step1: Apply translation \((x,y)\to(x - 4,y)\)

For point \(A(1,2)\): \(x=1,y = 2\), \(x-4=1 - 4=-3\), \(y = 2\), so \(A'(-3,2)\)
For point \(B(5,4)\): \(x = 5,y=4\), \(x-4=5 - 4 = 1\), \(y = 4\), so \(B'(1,4)\)
For point \(C(3,0)\): \(x = 3,y = 0\), \(x-4=3 - 4=-1\), \(y = 0\), so \(C'(-1,0)\)

Step2: Apply reflection in the \(x\) - axis \((x,y)\to(x,-y)\)

For \(A'(-3,2)\): \(x=-3,y = 2\), new \(y=-2\), so \(A''(-3,-2)\)
For \(B'(1,4)\): \(x = 1,y = 4\), new \(y=-4\), so \(B''(1,-4)\)
For \(C'(-1,0)\): \(x=-1,y = 0\), new \(y = 0\), so \(C''(-1,0)\)

Step1: Apply translation \(\langle0,-3

angle\) (i.e., \((x,y)\to(x,y - 3)\))
For \(A(1,2)\): \(x = 1,y=2\), \(y-3=2-3=-1\), so \(A'(1,-1)\)
For \(B(5,4)\): \(x = 5,y = 4\), \(y-3=4 - 3=1\), so \(B'(5,1)\)
For \(C(3,0)\): \(x = 3,y = 0\), \(y-3=0 - 3=-3\), so \(C'(3,-3)\)

Step2: Apply reflection in the \(y\) - axis \((x,y)\to(-x,y)\)

For \(A'(1,-1)\): \(x = 1,y=-1\), new \(x=-1\), so \(A''(-1,-1)\)
For \(B'(5,1)\): \(x = 5,y = 1\), new \(x=-5\), so \(B''(-5,1)\)
For \(C'(3,-3)\): \(x = 3,y=-3\), new \(x=-3\), so \(C''(-3,-3)\)

Step1: Apply translation \(\langle4,5

angle\) (i.e., \((x,y)\to(x + 4,y + 5)\))
For \(A(-3,-5)\): \(x=-3,y=-5\), \(x + 4=-3+4 = 1\), \(y + 5=-5 + 5=0\), so \(A'(1,0)\)
For \(B(0,-5)\): \(x = 0,y=-5\), \(x + 4=0+4 = 4\), \(y + 5=-5 + 5=0\), so \(B'(4,0)\)
For \(C(1,-2)\): \(x = 1,y=-2\), \(x + 4=1+4 = 5\), \(y + 5=-2 + 5=3\), so \(C'(5,3)\)

Step2: Apply reflection in the line \(y=x\) \((x,y)\to(y,x)\)

For \(A'(1,0)\): \(x = 1,y = 0\), new \(x = 0,y = 1\), so \(A''(0,1)\)
For \(B'(4,0)\): \(x = 4,y = 0\), new \(x = 0,y = 4\), so \(B''(0,4)\)
For \(C'(5,3)\): \(x = 5,y = 3\), new \(x = 3,y = 5\), so \(C''(3,5)\)

Answer:

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

Problem 2