Question

$\overline{fg}$ is a translation of $\overline{fg}$. write the translation rule.

Explanation:

Step1: Find coordinates of F, G, F', G'

From the graph:

• \( F(-2, -4) \), \( G(-5, -6) \)
• \( F'(7, 1) \), \( G'(5, -1) \)

Step2: Calculate horizontal (Δx) and vertical (Δy) shifts

For \( F \to F' \):
\( \Delta x = 7 - (-2) = 9 \)
\( \Delta y = 1 - (-4) = 5 \)

Check with \( G \to G' \):
\( \Delta x = 5 - (-5) = 10 \)? Wait, no—wait, recheck \( G \)'s coordinate. Wait, \( G \) is at \( (-5, -6) \)? Wait, no, looking at the grid: \( G \) is at \( x=-5 \), \( y=-6 \)? Wait, no, the blue \( G \) is at \( x=-5 \)? Wait, no, the grid lines: each square is 1 unit. Let's re-express:

Wait, \( F \): x=-2, y=-4 (blue dot). \( G \): x=-5, y=-6 (blue dot). \( F' \): x=7, y=1 (yellow dot). \( G' \): x=5, y=-1 (yellow dot). Wait, no, \( G' \) is at (5, -1)? Wait, the yellow \( G' \) is at x=5, y=-1? Wait, no, the graph: \( G' \) is at (5, -1)? Wait, no, looking at the yellow segment: \( G' \) is at (5, -1)? Wait, no, the x-axis: 5 is between 4 and 6. Wait, maybe I misread \( G \)'s x-coordinate. Wait, \( G \) is at x=-5? No, wait, the blue \( G \) is at x=-5? Wait, no, the grid: from -10 to 10, each line is 1. So \( G \) (blue) is at x=-5? Wait, no, the blue \( G \) is at x=-5? Wait, no, let's count: from y=-6, x=-5? Wait, no, the blue \( G \) is at ( -5, -6 )? And \( F \) is at ( -2, -4 ). Then \( F' \) is at (7, 1), \( G' \) is at (5, -1)? Wait, no, \( G' \) is at (5, -1)? Wait, no, the yellow \( G' \) is at x=5, y=-1? Wait, no, the yellow segment: \( G' \) is at (5, -1)? Wait, no, the x-coordinate of \( G' \): the yellow \( G' \) is at x=5? Wait, no, the grid: \( G' \) is at (5, -1)? Wait, no, let's recalculate \( \Delta x \) and \( \Delta y \) correctly.

Wait, \( F(-2, -4) \) to \( F'(7, 1) \):
\( \Delta x = 7 - (-2) = 9 \)
\( \Delta y = 1 - (-4) = 5 \)

\( G(-5, -6) \) to \( G'(5, -1) \):
\( \Delta x = 5 - (-5) = 10 \)? That's inconsistent. Wait, I must have misread \( G \)'s x-coordinate. Wait, no, \( G \) is at x=-5? Wait, no, the blue \( G \) is at x=-5? Wait, no, the blue \( G \) is at x=-5? Wait, no, let's look again: the blue segment \( FG \): \( F \) is at ( -2, -4 ), \( G \) is at ( -5, -6 )? No, that would make the vector from \( G \) to \( F \) as (3, 2). Then \( F' \) to \( G' \) should be (3, 2) as well. So \( F' \) is (7, 1), so \( G' \) should be (7 - 3, 1 - 2 ) = (4, -1). Ah! I misread \( G' \)'s x-coordinate. \( G' \) is at (4, -1), not (5, -1). That's the mistake. So \( G' \) is at (4, -1). Then:

\( G(-5, -6) \) to \( G'(4, -1) \):
\( \Delta x = 4 - (-5) = 9 \)
\( \Delta y = -1 - (-6) = 5 \)

Yes! That matches \( F \) to \( F' \): \( \Delta x = 7 - (-2) = 9 \), \( \Delta y = 1 - (-4) = 5 \). Perfect, so the translation is 9 units right (Δx=9) and 5 units up (Δy=5).

Answer:

The translation rule is \((x, y) \to (x + 9, y + 5)\)