Question

graph the image of △pqr after a translation 7 units left.

Explanation:

Step1: Find coordinates of P, Q, R

From the graph, \( P(4,7) \), \( Q(4,9) \), \( R(3,9) \) (assuming the grid and points are as per standard coordinate system, visually estimating the coordinates: P is at x=4, y=7; Q at x=4, y=9; R at x=3, y=9? Wait, no, looking again: Wait, the y-axis: R is at (3,9)? Wait, no, the grid: let's check the x and y. The x-axis: from -10 to 10, y from -10 to 10. Let's re-identify:

Looking at the triangle:

• Point P: x=4, y=7 (since it's 4 units right of origin on x, 7 up on y)
• Point Q: x=4, y=9 (same x as P, 2 up)
• Point R: x=3, y=9 (1 left of Q, same y)

Wait, no, maybe I misread. Wait, the yellow triangle: R is at (3,9)? Wait, no, the x-coordinate: let's count the grid lines. Each grid is 1 unit. So:

• R: x=3? Wait, no, the origin is (0,0). So moving right: x=1,2,3,4,... So R is at x=3, y=9? Q at x=4, y=8? Wait, maybe I made a mistake. Let's look again.

Wait, the y-axis: the top is 10. So R is at (3,9)? Q at (4,8)? P at (4,7)? Let's confirm:

Looking at the graph:

• P: (4,7) (x=4, y=7)
• Q: (4,8) (x=4, y=8)
• R: (3,9) (x=3, y=9)

Yes, that makes sense. So coordinates:

\( P(4,7) \), \( Q(4,8) \), \( R(3,9) \)

Step2: Apply translation 7 units left

Translation 7 units left means subtract 7 from the x-coordinate (since left is negative x-direction), y-coordinate remains same.

For \( P(4,7) \): new x = \( 4 - 7 = -3 \), y = 7 → \( P'(-3,7) \)

For \( Q(4,8) \): new x = \( 4 - 7 = -3 \), y = 8 → \( Q'(-3,8) \)

For \( R(3,9) \): new x = \( 3 - 7 = -4 \), y = 9 → \( R'(-4,9) \)

Step3: Plot the new points

Plot \( P'(-3,7) \), \( Q'(-3,8) \), \( R'(-4,9) \) and connect them to form the translated triangle.

Answer:

The translated triangle \( \triangle P'Q'R' \) has vertices at \( P'(-3,7) \), \( Q'(-3,8) \), \( R'(-4,9) \). (To graph, plot these points and draw the triangle.)