Question

he list of transformations applied to the square ABCD
, in sequence, is shown—

Rotate by 270∘
clockwise around the origin
Reflect across line y=x
Dilate by scale factor of 2
with origin as the center of dilation
Draw the resulting square on the coordinate plane. To draw a square, click on polygon and select all four vertices of the square in order, close the polygon by selecting the first vertex at the end.

Answer 1

To perform the transformations on square ABCD which we will assume has vertices at \( A(1, 1) \), \( B(1, -1) \), \( C(-1, -1) \), and \( D(-1, 1) \) (these points form a square centered at the origin with side length 2), we will apply each transformation step by step.

### Step 1: Rotate by 270° clockwise around the origin

**Rotation Formula**:
For a point \( (x, y) \), rotating it by \( \theta \) degrees results in:
\[
(x', y') = (x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta)
\]

For a rotation of \( 270^\circ \) clockwise, \( \cos(270^\circ) = 0 \) and \( \sin(270^\circ) = -1 \).
Thus:
\[
(x', y') = (y, -x)
\]

**Applying rotation to each vertex**:
- \( A(1, 1) \) becomes \( (1, -1) \)
- \( B(1, -1) \) becomes \( (-1, -1) \)
- \( C(-1, -1) \) becomes \( (-1, 1) \)
- \( D(-1, 1) \) becomes \( (1, 1) \)

After this step, the new coordinates are:
- \( A'(1, -1) \)
- \( B'(-1, -1) \)
- \( C'(-1, 1) \)
- \( D'(1, 1) \)

### Step 2: Reflect across the line \( y = x \)

**Reflection Formula**:
For a point \( (x, y) \), reflecting it across the line \( y = x \) results in:
\[
(x', y') = (y, x)
\]

**Applying reflection to each vertex**:
- \( A'(1, -1) \) becomes \( (-1, 1) \)
- \( B'(-1, -1) \) becomes \( (-1, -1) \)
- \( C'(-1, 1) \) becomes \( (1, -1) \)
- \( D'(1, 1) \) becomes \( (1, 1) \)

After this step, the new coordinates are:
- \( A''(-1, 1) \)
- \( B''(-1, -1) \)
- \( C''(1, -1) \)
- \( D''(1, 1) \)

### Step 3: Dilate by scale factor of 2 with origin as the center of dilation

**Dilation Formula**:
For a point \( (x, y) \), dilating it by a factor \( k \) results in:
\[
(x', y') = (kx, ky)
\]

**Applying dilation by a factor of 2 to each vertex**:
- \( A''(-1, 1) \) becomes \( (-2, 2) \)
- \( B''(-1, -1) \) becomes \( (-2, -2) \)
- \( C''(1, -1) \) becomes \( (2, -2) \)
- \( D''(1, 1) \) becomes \( (2, 2) \)

### Final coordinates:
- \( A'''(-2, 2) \)
- \( B'''(-2, -2) \)
- \( C'''(2, -2) \)
- \( D'''(2, 2) \)

### Drawing the resulting square

Now you can plot the vertices \((-2, 2)\), \((-2, -2)\), \((2, -2)\), and \((2, 2)\) on a coordinate plane and connect them as a polygon:

1. **Plot and Connect**:
- Start at \( (-2, 2) \)
- Draw a line to \( (-2, -2) \)
- Draw a line to \( (2, -2) \)
- Draw a line to \( (2, 2) \)
- Connect back to \( (-2, 2) \) to close the square.

This will give you the resulting dilated square on the coordinate plane after performing all the transformations!