Question

1. using the rule (x, y) → (-y, x) find the image points:

a: (2, 3)
b: (-5, 6)
c: (-1, -3)
a: ( , )
b: ( , )
c: ( , )

1. using the rule (x, y) → (x, -y) find the image points:

a: (2, 3)
b: (-5, 6)
c: (-1, -3)
a: ( , )
b: ( , )
c: ( , )

1. transform the pre-image using the rule (x, y) → (-x, -y). draw and label the image on the graph.

a. provide the coordinates for the pre-image and image.
t: ( , ) i: ( , )
i: ( , ) i: ( , )
e: ( , ) e: ( , )
t: ( , ) i: ( , )
b. what transformation was performed? be specific!

Explanation:

Problem 36: Using the rule \((x, y) \to (-y, x)\) to find image points

Let's take the given pre - image points (assuming the pre - image points for this problem are the same as in related problems, say \(A(2,3)\), \(B(-5,6)\), \(C(-1,-3)\))

Step 1: Find the image of point \(A(2,3)\)

For the transformation rule \((x,y)\to(-y,x)\), substitute \(x = 2\) and \(y=3\) into the rule.
The new \(x\) - coordinate is \(-y=-3\) and the new \(y\) - coordinate is \(x = 2\). So \(A'(-3,2)\)

Step 2: Find the image of point \(B(-5,6)\)

Substitute \(x=-5\) and \(y = 6\) into the rule \((x,y)\to(-y,x)\).
The new \(x\) - coordinate is \(-y=-6\) and the new \(y\) - coordinate is \(x=-5\). So \(B'(-6,-5)\)

Step 3: Find the image of point \(C(-1,-3)\)

Substitute \(x = - 1\) and \(y=-3\) into the rule \((x,y)\to(-y,x)\).
The new \(x\) - coordinate is \(-y = 3\) and the new \(y\) - coordinate is \(x=-1\). So \(C'(3,-1)\)

Problem 37: Using the rule \((x, y) \to (x,-y)\) to find image points

Step 1: Find the image of point \(A(2,3)\)

For the transformation rule \((x,y)\to(x,-y)\), substitute \(x = 2\) and \(y = 3\).
The new \(x\) - coordinate is \(x = 2\) and the new \(y\) - coordinate is \(-y=-3\). So \(A'(2,-3)\)

Step 2: Find the image of point \(B(-5,6)\)

Substitute \(x=-5\) and \(y = 6\) into the rule \((x,y)\to(x,-y)\).
The new \(x\) - coordinate is \(x=-5\) and the new \(y\) - coordinate is \(-y=-6\). So \(B'(-5,-6)\)

Step 3: Find the image of point \(C(-1,-3)\)

Substitute \(x=-1\) and \(y=-3\) into the rule \((x,y)\to(x,-y)\).
The new \(x\) - coordinate is \(x=-1\) and the new \(y\) - coordinate is \(-y = 3\). So \(C'(-1,3)\)

Problem 38a: Provide coordinates for pre - image and image (assuming the pre - image has points \(T\), \(I\), \(E\) from the graph)

Looking at the graph (assuming a typical grid - based graph for a triangle or a figure):
Let's assume the pre - image coordinates:

• Let \(T\) be at \((0,2)\) (on the \(y\) - axis, 2 units up), \(I\) be at \((0,0)\) (origin), \(E\) be at \((2,0)\) (on the \(x\) - axis, 2 units right)

For the transformation rule \((x,y)\to(-x,-y)\):

Step 1: Find the image of \(T(0,2)\)

Substitute \(x = 0\) and \(y = 2\) into \((x,y)\to(-x,-y)\).
New \(x=-0 = 0\), new \(y=-2\). So \(T'(0,-2)\)

Step 2: Find the image of \(I(0,0)\)

Substitute \(x = 0\) and \(y = 0\) into \((x,y)\to(-x,-y)\).
New \(x=-0 = 0\), new \(y=-0 = 0\). So \(I'(0,0)\)

Step 3: Find the image of \(E(2,0)\)

Substitute \(x = 2\) and \(y = 0\) into \((x,y)\to(-x,-y)\).
New \(x=-2\), new \(y=-0 = 0\). So \(E'(-2,0)\)

Problem 38b: What transformation was performed?

The rule \((x,y)\to(-x,-y)\) represents a rotation of \(180^{\circ}\) about the origin. In a \(180^{\circ}\) rotation about the origin, each point \((x,y)\) is mapped to \((-x,-y)\) because the rotation matrix for \(180^{\circ}\) is \(

$$\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$$

\), and when we multiply the vector \(

$$\begin{pmatrix}x\\y\end{pmatrix}$$

\) by this matrix, we get \(

$$\begin{pmatrix}-x\\-y\end{pmatrix}$$

\)

Final Answers

Problem 36

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

Problem 37

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

Problem 38a

Pre - image: \(T(0,2)\), \(I(0,0)\), \(E(2,0)\); Image: \(T'(0,-2)\), \(I'(0,0)\), \(E'(-2,0)\) (answers may vary slightly depending on the exact graph coordinates)

Problem 38b

Rotation of \(180^{\circ}\) about the origin

Answer:

Problem 36: Using the rule \((x, y) \to (-y, x)\) to find image points

Let's take the given pre - image points (assuming the pre - image points for this problem are the same as in related problems, say \(A(2,3)\), \(B(-5,6)\), \(C(-1,-3)\))

Step 1: Find the image of point \(A(2,3)\)

For the transformation rule \((x,y)\to(-y,x)\), substitute \(x = 2\) and \(y=3\) into the rule.
The new \(x\) - coordinate is \(-y=-3\) and the new \(y\) - coordinate is \(x = 2\). So \(A'(-3,2)\)

Step 2: Find the image of point \(B(-5,6)\)

Substitute \(x=-5\) and \(y = 6\) into the rule \((x,y)\to(-y,x)\).
The new \(x\) - coordinate is \(-y=-6\) and the new \(y\) - coordinate is \(x=-5\). So \(B'(-6,-5)\)

Step 3: Find the image of point \(C(-1,-3)\)

Substitute \(x = - 1\) and \(y=-3\) into the rule \((x,y)\to(-y,x)\).
The new \(x\) - coordinate is \(-y = 3\) and the new \(y\) - coordinate is \(x=-1\). So \(C'(3,-1)\)

Problem 37: Using the rule \((x, y) \to (x,-y)\) to find image points

Step 1: Find the image of point \(A(2,3)\)

For the transformation rule \((x,y)\to(x,-y)\), substitute \(x = 2\) and \(y = 3\).
The new \(x\) - coordinate is \(x = 2\) and the new \(y\) - coordinate is \(-y=-3\). So \(A'(2,-3)\)

Step 2: Find the image of point \(B(-5,6)\)

Substitute \(x=-5\) and \(y = 6\) into the rule \((x,y)\to(x,-y)\).
The new \(x\) - coordinate is \(x=-5\) and the new \(y\) - coordinate is \(-y=-6\). So \(B'(-5,-6)\)

Step 3: Find the image of point \(C(-1,-3)\)

Substitute \(x=-1\) and \(y=-3\) into the rule \((x,y)\to(x,-y)\).
The new \(x\) - coordinate is \(x=-1\) and the new \(y\) - coordinate is \(-y = 3\). So \(C'(-1,3)\)

Problem 38a: Provide coordinates for pre - image and image (assuming the pre - image has points \(T\), \(I\), \(E\) from the graph)

Looking at the graph (assuming a typical grid - based graph for a triangle or a figure):
Let's assume the pre - image coordinates:

• Let \(T\) be at \((0,2)\) (on the \(y\) - axis, 2 units up), \(I\) be at \((0,0)\) (origin), \(E\) be at \((2,0)\) (on the \(x\) - axis, 2 units right)

For the transformation rule \((x,y)\to(-x,-y)\):

Step 1: Find the image of \(T(0,2)\)

Substitute \(x = 0\) and \(y = 2\) into \((x,y)\to(-x,-y)\).
New \(x=-0 = 0\), new \(y=-2\). So \(T'(0,-2)\)

Step 2: Find the image of \(I(0,0)\)

Substitute \(x = 0\) and \(y = 0\) into \((x,y)\to(-x,-y)\).
New \(x=-0 = 0\), new \(y=-0 = 0\). So \(I'(0,0)\)

Step 3: Find the image of \(E(2,0)\)

Substitute \(x = 2\) and \(y = 0\) into \((x,y)\to(-x,-y)\).
New \(x=-2\), new \(y=-0 = 0\). So \(E'(-2,0)\)

Problem 38b: What transformation was performed?

The rule \((x,y)\to(-x,-y)\) represents a rotation of \(180^{\circ}\) about the origin. In a \(180^{\circ}\) rotation about the origin, each point \((x,y)\) is mapped to \((-x,-y)\) because the rotation matrix for \(180^{\circ}\) is \(

$$\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$$

\), and when we multiply the vector \(

$$\begin{pmatrix}x\\y\end{pmatrix}$$

\) by this matrix, we get \(

$$\begin{pmatrix}-x\\-y\end{pmatrix}$$

\)

Final Answers

Problem 36

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

Problem 37

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

Problem 38a

Pre - image: \(T(0,2)\), \(I(0,0)\), \(E(2,0)\); Image: \(T'(0,-2)\), \(I'(0,0)\), \(E'(-2,0)\) (answers may vary slightly depending on the exact graph coordinates)

Problem 38b

Rotation of \(180^{\circ}\) about the origin