Question

which vector describes the translation.

Explanation:

Step1: Identify coordinates of a point (e.g., V)

Let's take point \( V \) from the original figure. From the grid, \( V \) is at \( (0, 3) \) (assuming each grid square is 1 unit). The translated point \( V' \) is at \( (-3, 0) \).

Step2: Calculate the translation vector

The translation vector \( \vec{v} = (x' - x, y' - y) \), where \( (x, y) \) is the original coordinate and \( (x', y') \) is the translated coordinate. For \( V \) and \( V' \): \( x' - x = -3 - 0 = -3 \), \( y' - y = 0 - 3 = -3 \). Wait, maybe better to check another point. Let's take \( W \): Original \( W \) (let's assume coordinates: from \( V(0,3) \), \( W \) is, say, \( (4, 4) \)? Wait, maybe my initial grid assumption is off. Wait, looking at the blue triangle (translated) and black triangle (original). Let's find the horizontal and vertical change. Let's take point \( V \): original \( V \) is on the y-axis, say \( (0, 3) \), translated \( V' \) is at \( (-3, 0) \). Wait, horizontal change: from \( x=0 \) to \( x=-3 \): change of \( -3 \). Vertical change: from \( y=3 \) to \( y=0 \): change of \( -3 \)? Wait, no, maybe I messed up. Wait, let's look at the y-coordinate: original triangle is above the x-axis, translated is on or below? Wait, the blue triangle (translated) has \( V' \) on the x-axis (y=0), original \( V \) is at y=3 (since it's 3 units above x-axis). So vertical change: \( 0 - 3 = -3 \). Horizontal change: original \( V \) is at x=0, translated \( V' \) is at x=-3 (3 units left). So the translation vector is \( \langle -3, -3
angle \)? Wait, no, maybe check point \( X \). Original \( X \): let's say original \( X \) is at \( (3, 2) \) (since from \( V(0,3) \), moving right 3, down 1). Translated \( X' \) is at \( (0, -1) \)? Wait, no, the blue triangle: \( X' \) is at (3, -1)? Wait, maybe I need to count the grid squares. Let's assume each grid is 1 unit. Let's find the coordinates properly.

Original \( V \): (0, 3) (on y-axis, 3 units up). Translated \( V' \): (-3, 0) (3 units left, 3 units down). So the translation vector is \( (x_{\text{translated}} - x_{\text{original}}, y_{\text{translated}} - y_{\text{original}}) = (-3 - 0, 0 - 3) = (-3, -3) \). Wait, but maybe the vector is \( \langle -3, -3
angle \), meaning 3 units left and 3 units down.

Wait, maybe another approach: translation vector is the change from original to image. So for any point \( P(x, y) \) to \( P'(x + a, y + b) \), the vector is \( \langle a, b
angle \). So if \( V(0, 3) \) becomes \( V'(-3, 0) \), then \( 0 + a = -3 \implies a = -3 \), \( 3 + b = 0 \implies b = -3 \). So the vector is \( \langle -3, -3
angle \).

Answer:

The translation vector is \( \langle -3, -3
angle \) (or written as \( (-3, -3) \) depending on notation, representing 3 units left and 3 units down).