Question

write the coordinates of the vertices after a translation 8 units left.
b (□, )
c ( , )
d ( , )
e ( , )

Explanation:

Step1: Find original coordinates

First, identify the original coordinates of each vertex from the graph:

• \( B \): Looking at the grid, \( B \) is at \( (6, -7) \)
• \( C \): \( C \) is at \( (6, -6) \)
• \( D \): \( D \) is at \( (7, -6) \) (Wait, no, let's check again. Wait, the square: let's see the x and y. Wait, the grid: x-axis, y-axis. Let's re-express:

Wait, the y-coordinate for B: looking at the vertical line, B is at y = -7? Wait, no, the grid lines: the bottom part, the y-axis has -6, -7, -8? Wait, no, the graph: the square is at x around 6,7 and y around -6, -7. Wait, let's check again.

Wait, the original coordinates:

• \( B \): x=6, y=-7? Wait, no, the vertical lines: each grid is 1 unit. Let's see, the point B: when x=6, y=-7? Wait, no, the y-axis: the lines are at y= -6, -7, -8? Wait, the square has B, C, D, E. Let's see:

From the graph, let's find each point:

• \( B \): (6, -7)
• \( C \): (6, -6)
• \( D \): (7, -6)
• \( E \): (7, -7)

Wait, no, maybe I made a mistake. Wait, the square: B to C is up 1, C to D is right 1, D to E is down 1, E to B is left 1. So:

• \( B \): (6, -7)
• \( C \): (6, -6) (up 1 from B)
• \( D \): (7, -6) (right 1 from C)
• \( E \): (7, -7) (down 1 from D)

Yes, that makes a square.

Step2: Apply translation (8 units left)

A translation 8 units left means we subtract 8 from the x-coordinate (since left is negative x-direction), and y-coordinate remains the same.

The translation rule for 8 units left is \( (x, y) \to (x - 8, y) \)

For \( B(6, -7) \):

New x: \( 6 - 8 = -2 \), y remains -7. So \( B'(-2, -7) \)

For \( C(6, -6) \):

New x: \( 6 - 8 = -2 \), y remains -6. So \( C'(-2, -6) \)

For \( D(7, -6) \):

New x: \( 7 - 8 = -1 \), y remains -6. So \( D'(-1, -6) \)

For \( E(7, -7) \):

New x: \( 7 - 8 = -1 \), y remains -7. So \( E'(-1, -7) \)

Wait, wait, maybe I messed up the original x-coordinates. Wait, let's check the original x: the square is at x=6 and 7? Wait, no, looking at the grid, the x-axis: from -10 to 10, each grid is 1. So the original x for B: is it 6? Wait, the x=6 is at the vertical line labeled 6. So yes, B is at (6, -7), C at (6, -6), D at (7, -6), E at (7, -7). Then translating 8 units left:

• \( B' \): \( 6 - 8 = -2 \), so \( (-2, -7) \)
• \( C' \): \( 6 - 8 = -2 \), so \( (-2, -6) \)
• \( D' \): \( 7 - 8 = -1 \), so \( (-1, -6) \)
• \( E' \): \( 7 - 8 = -1 \), so \( (-1, -7) \)

Wait, but let's confirm the original coordinates again. Maybe I made a mistake in the original y. Let's see the y-axis: the lines are at y= -6, -7, -8? Wait, the square is below the x-axis (y=0). So B is at y=-7? Wait, the grid lines: the horizontal lines (y-axis) have -6, -7, -8. So the square is between y=-6 and y=-7, and x=6 and x=7. So yes, original coordinates:

• \( B(6, -7) \)
• \( C(6, -6) \)
• \( D(7, -6) \)
• \( E(7, -7) \)

Then translation 8 units left: subtract 8 from x, y remains.

So:

• \( B' \): \( (6 - 8, -7) = (-2, -7) \)
• \( C' \): \( (6 - 8, -6) = (-2, -6) \)
• \( D' \): \( (7 - 8, -6) = (-1, -6) \)
• \( E' \): \( (7 - 8, -7) = (-1, -7) \)

Wait, but maybe the original x was 6 and 7? Wait, let's check the grid again. The x-axis: from 0 to 10 on the right. So 6 is 6 units right of 0, 7 is 7 units right. Translating 8 units left: 6 - 8 = -2, 7 - 8 = -1. Y-coordinates remain -7 and -6.

Yes, that makes sense.

Answer:

• \( B'(-2, -7) \)
• \( C'(-2, -6) \)
• \( D'(-1, -6) \)
• \( E'(-1, -7) \)