Question

write a rule for the translation of △lmn to △l′m′n′.
(x, y) → (□, □)

Explanation:

Step1: Identify coordinates of a point

Let's take point \( L \) from \( \triangle LMN \) and \( L' \) from \( \triangle L'M'N' \). From the graph, \( L \) has coordinates \( (1, -1) \) (wait, actually looking at the grid, let's recheck. Wait, \( L \) is at \( (1, -1) \)? Wait no, looking at the blue triangle: \( L \) is at \( (1, -1) \)? Wait maybe better to take \( N \): \( N \) is at \( (5, -1) \), \( N' \) is at \( (0, 1) \)? Wait no, maybe I misread. Wait the blue triangle: \( L \) is at \( (1, -1) \)? Wait no, let's look at the grid. The x-axis: from -4 to 6, y-axis from -2 to 4. Let's take point \( M \): blue \( M \) is at \( (4, 0) \)? Wait red \( M' \) is at \( (-2, 3) \)? Wait maybe better to take \( L \): blue \( L \) is at \( (1, -1) \), red \( L' \) is at \( (-4, 1) \)? Wait no, maybe I made a mistake. Wait the blue triangle: \( L \) is at \( (1, -1) \), \( M \) at \( (4, 0) \), \( N \) at \( (5, -1) \). Red triangle: \( L' \) at \( (-4, 1) \), \( M' \) at \( (-1, 3) \), \( N' \) at \( (0, 1) \). Wait no, maybe the blue \( L \) is at \( (1, -1) \), red \( L' \) at \( (-4, 1) \). So the change in x: \( -4 - 1 = -5 \), change in y: \( 1 - (-1) = 2 \). Wait or maybe another point. Wait let's take \( N \): blue \( N \) is at \( (5, -1) \), red \( N' \) is at \( (0, 1) \). So \( x \)-change: \( 0 - 5 = -5 \), \( y \)-change: \( 1 - (-1) = 2 \). Yes, that's consistent. So the translation is 5 units left (x decreases by 5) and 2 units up (y increases by 2).

Step2: Write the translation rule

For a point \( (x, y) \), after translation, the new x-coordinate is \( x - 5 \) and new y-coordinate is \( y + 2 \). So the rule is \( (x, y) \to (x - 5, y + 2) \).

Answer:

\( (x - 5, y + 2) \)