Question

1. naveen plotted his triangular garden on a coordinate plane. what are the vertices of the image of his garden if it is reflected in the line ( y = x )?
2. the image of ( a(-1, 1) ) after a reflection is ( a(-1, -3) ). which reflection in a vertical or horizontal line produces the image of ( a )?

Explanation:

Problem 12 (assuming we first identify the original vertices from the graph, let's assume the original triangle has vertices, for example, let's say from the grid: let's suppose the original vertices are, say, \( (-4, 4) \), \( (-5, 1) \), and \( (4, 1) \) (we need to get the actual coordinates from the grid, but since the grid is given, let's proceed with the reflection rule over \( y = x \), which is \( (x, y) \to (y, x) \))

Step 1: Recall the reflection rule over \( y = x \)

The rule for reflecting a point \( (x, y) \) over the line \( y = x \) is to swap the \( x \)-coordinate and the \( y \)-coordinate, so the image of \( (x, y) \) is \( (y, x) \).

Step 2: Identify original vertices (from the graph, let's assume the three vertices are, for example, let's look at the grid:

• Let's say the left - most bottom vertex: \( (-5, 1) \)
• The top - left vertex: \( (-4, 4) \)
• The right - most bottom vertex: \( (4, 1) \)

Step 3: Apply the reflection rule to each vertex

• For \( (-5, 1) \): Swap \( x \) and \( y \), we get \( (1, -5) \)
• For \( (-4, 4) \): Swap \( x \) and \( y \), we get \( (4, -4) \)
• For \( (4, 1) \): Swap \( x \) and \( y \), we get \( (1, 4) \)

(Note: The actual coordinates depend on the grid. Let's correct the identification. Let's re - examine the grid:

Looking at the graph, let's assume the three vertices are:

Vertex 1: Let's say the left vertex is at \( (-5, 1) \) (x = - 5, y = 1)

Vertex 2: The top vertex is at \( (-4, 4) \) (x=-4, y = 4)

Vertex 3: The right vertex is at \( (4, 1) \) (x = 4, y = 1)

After reflection over \( y=x \):

Vertex 1: \( (1, -5) \)

Vertex 2: \( (4, -4) \)

Vertex 3: \( (1, 4) \)

But if the original vertices are different, we need to adjust. Let's take a more accurate approach. Let's suppose from the grid:

Let the three vertices be \( A(-5,1) \), \( B(-4,4) \), \( C(4,1) \)

Reflection over \( y = x \): \( (x,y)\to(y,x) \)

So \( A'=(1, - 5) \), \( B'=(4, - 4) \), \( C'=(1, 4) \)

Problem 13

Step 1: Analyze the coordinates of \( A \) and \( A' \)

The coordinates of \( A \) are \( (-1,1) \) and the coordinates of \( A' \) are \( (-1, - 3) \).

Step 2: Check the type of reflection (horizontal or vertical line)

• For a horizontal line reflection (\( y = k \)), the \( x \)-coordinate remains the same, and the \( y \)-coordinate changes. The formula for reflection over \( y=k \) is \( (x,y)\to(x,2k - y) \)
• For a vertical line reflection (\( x = h \)), the \( y \)-coordinate remains the same, and the \( x \)-coordinate changes. The formula for reflection over \( x = h \) is \( (x,y)\to(2h - x,y) \)

Since the \( x \)-coordinate of \( A \) and \( A' \) is the same (\( x=-1 \)), it is a reflection over a horizontal line.

Step 3: Find the equation of the horizontal line

We know that for a reflection over \( y = k \), \( y'=2k - y \). We have \( y = 1 \) and \( y'=-3 \)

So, \( - 3=2k - 1 \)

Step 4: Solve for \( k \)

Add 1 to both sides: \( -3 + 1=2k\)

\( -2 = 2k\)

Divide both sides by 2: \( k=-1 \)

So the reflection is over the horizontal line \( y=-1 \)

Answer:

s:

Problem 12 (assuming the original vertices are \( (-5,1) \), \( (-4,4) \), \( (4,1) \))

The vertices of the reflected triangle are \( (1, - 5) \), \( (4, - 4) \), and \( (1, 4) \) (the actual answer depends on the correct original coordinates from the grid. If the original vertices are, for example, let's say the correct original vertices from the graph are \( (-5,1) \), \( (-4,4) \), \( (4,1) \), then the reflected vertices are as above. If the original vertices are different, recalculate using \( (x,y)\to(y,x) \))

Problem 13

The reflection is over the horizontal line \( y = - 1 \)