Question

1). translate the quadrilateral abcd - 2 horizontally and + 3 vertically. write new coordinates.
2). reflect the figure over y - axis. write new coordinates.

Explanation:

Step1: Find coordinates of ABCD in first problem

For parallelogram \(ABCD\) in the first graph:

• Point \(A\): \(x = - 1,y = 4\), so \(A(-1,4)\)
• Point \(B\): \(x = 5,y = 4\), so \(B(5,4)\)
• Point \(C\): \(x = 2,y = 2\), so \(C(2,2)\)
• Point \(D\): \(x=-4,y = 2\), so \(D(-4,2)\)

Step2: Translate coordinates

The translation rule is \((x,y)\to(x - 2,y + 3)\)

• For \(A(-1,4)\): \(x=-1-2=-3,y = 4 + 3=7\), so \(A'(-3,7)\)
• For \(B(5,4)\): \(x=5-2 = 3,y=4 + 3=7\), so \(B'(3,7)\)
• For \(C(2,2)\): \(x=2-2=0,y=2 + 3=5\), so \(C'(0,5)\)
• For \(D(-4,2)\): \(x=-4-2=-6,y=2 + 3=5\), so \(D'(-6,5)\)

Step3: Find coordinates of ABCD in second problem

For quadrilateral \(ABCD\) in the second graph:

• Point \(A\): \(x = 6,y = 3\), so \(A(6,3)\)
• Point \(B\): \(x=-5,y=-4\), so \(B(-5,-4)\)
• Point \(C\): \(x=-6,y = 2\), so \(C(-6,2)\)
• Point \(D\): \(x=-1,y = 6\), so \(D(-1,6)\)

Step4: Reflect coordinates over \(y\) - axis

The reflection rule over \(y\) - axis is \((x,y)\to(-x,y)\)

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

Answer:

1.

Parallelogram \(ABCD\)	Parallelogram \(A'B'C'D'\)
\(B(5,4)\)	\(B'(3,7)\)
\(C(2,2)\)	\(C'(0,5)\)
\(D(-4,2)\)	\(D'(-6,5)\)

2.

Quadrilateral \(ABCD\)	Quadrilateral \(A'B'C'D'\)
\(B(-5,-4)\)	\(B'(5,-4)\)
\(C(-6,2)\)	\(C'(6,2)\)
\(D(-1,6)\)	\(D'(1,6)\)