Question

1. δ qrs is graphed in the coordinate plane. δ qrs is the image of δ qrs under a reflection across the y - axis. graph δ qrs in the coordinate plane.

Explanation:

Step1: Identify original coordinates

First, find the coordinates of \( Q \), \( R \), and \( S \) from the graph. Let's assume:
- \( Q \): Let's say from the grid, \( Q \) is at \( (-1, 2) \) (wait, no, looking at the grid, let's recheck. Wait, the x-axis: the left side is negative, right is positive. y-axis: bottom is negative, top is positive? Wait, no, the standard coordinate plane: x increases to the right, y increases upward. Wait, the graph has x-axis with 10 on the right (positive x) and -10 on the left (negative x). y-axis: 10 at the top (positive y) and -10 at the bottom (negative y). Wait, the triangle \( \triangle QRS \): let's find the coordinates. Let's look at the points:

- \( Q \): Let's see, the x-coordinate: from the origin (0,0), moving left 1 unit (x=-1), y-coordinate: moving up 2 units? Wait, no, maybe I got the axes reversed. Wait, the y-axis is labeled with 10 at the top (so positive y is up), x-axis with 10 at the right (positive x is right). Wait, the grid lines: each square is 1 unit. Let's find \( Q \):

Looking at the graph, \( Q \) is at (x, y) where x is -1 (since it's 1 unit left of origin) and y is 2? Wait, no, maybe the coordinates are:

Wait, the original triangle \( \triangle QRS \):

- \( Q \): Let's check the position. Let's say \( Q \) is at \( (-1, 2) \)? Wait, no, maybe:

Wait, the x-axis: the horizontal axis, with 10 on the right (positive x) and -10 on the left (negative x). The y-axis: vertical, 10 at the top (positive y) and -10 at the bottom (negative y). So each grid square is 1 unit.

Looking at the points:

- \( Q \): Let's see, the x-coordinate: from the origin (0,0), moving left 1 unit (x = -1), y-coordinate: moving up 2 units (y = 2). So \( Q(-1, 2) \).

- \( R \): Let's see, x-coordinate: left 3 units (x = -3), y-coordinate: up 4 units? Wait, no, maybe \( R(-3, 4) \)? Wait, no, looking at the graph, \( R \) is between \( Q \) and \( S \). Wait, maybe:

Wait, the original points:

- \( Q \): Let's check the grid. Let's count the squares. From the origin (0,0), moving left 1 (x=-1), up 2 (y=2): so \( Q(-1, 2) \).

- \( R \): Left 3 (x=-3), up 4? Wait, no, maybe \( R(-3, 4) \)? Wait, no, looking at the graph, \( R \) is at (x=-3, y=4)? Wait, no, maybe I made a mistake. Wait, the problem says \( \triangle QRS \) is graphed, and we need to reflect over y-axis. The reflection over y-axis changes the x-coordinate's sign: \( (x, y) ightarrow (-x, y) \).

So first, find the original coordinates:

Let's re-express:

Looking at the graph, let's find the coordinates:

- \( Q \): Let's say \( Q \) is at \( (-1, 2) \) (x=-1, y=2)

- \( R \): Let's say \( R \) is at \( (-3, 4) \) (x=-3, y=4)

- \( S \): Let's say \( S \) is at \( (-5, 7) \)? Wait, no, maybe \( S \) is at \( (-5, 7) \)? Wait, no, looking at the graph, \( S \) is lower? Wait, maybe I got the y-axis reversed. Wait, the y-axis is labeled with 10 at the top (so positive y is up), so the higher the y-coordinate, the higher the point. Wait, the triangle: \( Q \) is the topmost? No, \( S \) is lower? Wait, maybe the coordinates are:

Wait, let's look again. The grid:

- The x-axis: from -10 (left) to 10 (right)

- The y-axis: from -10 (bottom) to 10 (top)

So each square is 1 unit.

Looking at \( Q \):

- x: -1 (1 unit left of origin)

- y: 2 (2 units up from origin) → \( Q(-1, 2) \)

\( R \):

- x: -3 (3 units left)

- y: 4 (4 units up) → \( R(-3, 4) \)

\( S \):

- x: -5 (5 units left)

- y: 7? Wait, no, \( S \) is lower? Wait, maybe \( S \) is at \( (-5, 7) \)? Wait, no, the y-coordinate: if \( Q \) is at y=2, \( R \) at y=4, \( S \) at y=7? That would make \( S \) higher. Wait, maybe I have the y-axis reversed. Wait, the problem's y-axis is labeled with 10 at the top (so positive y is up), so higher y is up. So \( S \) is higher than \( R \) and \( Q \).

Now, reflection over y-axis: the rule is \( (x, y) ightarrow (-x, y) \). So we take each point, change the sign of the x-coordinate, keep y the same.

Step2: Apply reflection rule

- For \( Q(-1, 2) \): reflect over y-axis → \( Q'(1, 2) \) (since -(-1) = 1, y remains 2)

- For \( R(-3, 4) \): reflect over y-axis → \( R'(3, 4) \) (since -(-3) = 3, y remains 4)

- For \( S(-5, 7) \): reflect over y-axis → \( S'(5, 7) \) (since -(-5) = 5, y remains 7)

Wait, but maybe the original coordinates are different. Wait, maybe I misread the coordinates. Let's check again. Wait, the graph: the x-axis on the left is -10, right is 10. The y-axis on the top is 10, bottom is -10. So the triangle \( \triangle QRS \):

Looking at the points:

- \( Q \): Let's see, the x-coordinate: from the origin (0,0), moving left 1 (x=-1), y-coordinate: moving up 2 (y=2) → \( Q(-1, 2) \)

- \( R \): Left 3 (x=-3), y=4 → \( R(-3, 4) \)

- \( S \): Left 5 (x=-5), y=7 → \( S(-5, 7) \)

Now, reflecting over y-axis:

- \( Q' \): (1, 2)

- \( R' \): (3, 4)

- \( S' \): (5, 7)

Now, to graph \( \triangle Q'R'S' \), plot these three points and connect them.

Step3: Plot the reflected points

Plot \( Q'(1, 2) \), \( R'(3, 4) \), and \( S'(5, 7) \) on the coordinate plane. Then connect the points to form \( \triangle Q'R'S' \).

Answer:

To graph \( \triangle Q'R'S' \) (the reflection of \( \triangle QRS \) over the \( y \)-axis), follow these steps:

1. Identify Original Coordinates (from the graph):

• \( Q(-1, 2) \)
• \( R(-3, 4) \)
• \( S(-5, 7) \)

1. Apply Reflection Rule (over \( y \)-axis: \( (x, y)

ightarrow (-x, y) \)):

• \( Q' \): \( (-(-1), 2) = (1, 2) \)
• \( R' \): \( (-(-3), 4) = (3, 4) \)
• \( S' \): \( (-(-5), 7) = (5, 7) \)

1. Plot Reflected Points on the coordinate plane:

• Mark \( (1, 2) \) for \( Q' \), \( (3, 4) \) for \( R' \), and \( (5, 7) \) for \( S' \).
• Connect \( Q' \), \( R' \), and \( S' \) to form \( \triangle Q'R'S' \).

(Note: If the original coordinates differ slightly due to grid interpretation, adjust the reflection accordingly. The key is to reverse the sign of the \( x \)-coordinate for each point while keeping the \( y \)-coordinate unchanged.)