Question

draw the image of quadrilateral abcd under a translation by 2 units to the right and 5 units down.

Explanation:

Step1: Identify coordinates of vertices

First, find the coordinates of \( A \), \( B \), \( C \), \( D \). From the graph:

• \( A(-4, 1) \)
• \( B(3, -1) \) (Wait, no, looking at the grid, let's recheck. Wait, the blue and green points: Let's correctly identify. Let's see the green points: \( A(-4, 1) \), \( D(-5, 3) \), \( C(4, 1) \), \( B(3, -1) \)? Wait, maybe better to get exact coordinates. Let's list each vertex:

Looking at the grid:

• \( A \): x=-4, y=1 (so \( A(-4, 1) \))
• \( B \): x=3, y=-1 ( \( B(3, -1) \))
• \( C \): x=4, y=1 ( \( C(4, 1) \))
• \( D \): x=-5, y=3 ( \( D(-5, 3) \))

Step2: Apply translation rule

Translation: 2 units right (add 2 to x-coordinate) and 5 units down (subtract 5 from y-coordinate). The translation rule is \( (x, y) \to (x + 2, y - 5) \).

For \( A(-4, 1) \):

New x: \( -4 + 2 = -2 \)
New y: \( 1 - 5 = -4 \)
So \( A'(-2, -4) \)

For \( B(3, -1) \):

New x: \( 3 + 2 = 5 \)
New y: \( -1 - 5 = -6 \)
So \( B'(5, -6) \)

For \( C(4, 1) \):

New x: \( 4 + 2 = 6 \)
New y: \( 1 - 5 = -4 \)
So \( C'(6, -4) \)

For \( D(-5, 3) \):

New x: \( -5 + 2 = -3 \)
New y: \( 3 - 5 = -2 \)
So \( D'(-3, -2) \)

Step3: Plot the new points and draw the quadrilateral

Plot \( A'(-2, -4) \), \( B'(5, -6) \), \( C'(6, -4) \), \( D'(-3, -2) \) on the grid and connect them in order to get the translated quadrilateral \( A'B'C'D' \).

Answer:

The translated quadrilateral has vertices at \( A'(-2, -4) \), \( B'(5, -6) \), \( C'(6, -4) \), \( D'(-3, -2) \). Plot these points and connect them to see the image. (Note: Since drawing is required, the key is applying the translation rule to each vertex's coordinates as shown above.)