Question

exit ticket seal the deal!
quadrilateral abcd has coordinates a(0,2), b(3,4), c(4,2),
and d(3,0). determine the coordinates of the vertices of
abcd after reflections in:

1. the y - axis \i know the rule for reflections

over the y - axis is (x,y)→(-x,y)\
a( , ) b( , ) c( , ) d( , )

1. the line y = 1 \i can use the graph to help me

find the coordinates of the image\
a( , ) b( , ) c( , ) d( , )

1. the line y = x \i know the rule for reflections

over y = x is (x,y)→(y,x)\
a( , ) b( , ) c( , ) d( , )

Explanation:

1. Reflection over the \(y\) - axis

Step1: Apply the rule \((x,y)\to(-x,y)\)

For point \(A(0,2)\):
\(x = 0,y = 2\), then \(A'(-0,2)=(0,2)\)
For point \(B(3,4)\):
\(x = 3,y = 4\), then \(B'(-3,4)\)
For point \(C(4,2)\):
\(x = 4,y = 2\), then \(C'(-4,2)\)
For point \(D(3,0)\):
\(x = 3,y = 0\), then \(D'(-3,0)\)

2. Reflection over the line \(y = 1\)

Step1: Use the formula for reflection over the line \(y = k\). The formula for reflecting a point \((x,y)\) over the line \(y=k\) is \((x,2k - y)\)

Here \(k = 1\), so the formula is \((x,2\times1 - y)=(x,2 - y)\)
For point \(A(0,2)\):
\(x = 0,y = 2\), then \(A'(0,2-2)=(0,0)\)
For point \(B(3,4)\):
\(x = 3,y = 4\), then \(B'(3,2 - 4)=(3,-2)\)
For point \(C(4,2)\):
\(x = 4,y = 2\), then \(C'(4,2-2)=(4,0)\)
For point \(D(3,0)\):
\(x = 3,y = 0\), then \(D'(3,2-0)=(3,2)\)

3. Reflection over the line \(y=x\)

Step1: Apply the rule \((x,y)\to(y,x)\)

For point \(A(0,2)\):
\(x = 0,y = 2\), then \(A'(2,0)\)
For point \(B(3,4)\):
\(x = 3,y = 4\), then \(B'(4,3)\)
For point \(C(4,2)\):
\(x = 4,y = 2\), then \(C'(2,4)\)
For point \(D(3,0)\):
\(x = 3,y = 0\), then \(D'(0,3)\)

Answer:

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