Question

name date ① consider △abc. a translate △abc 3 units right. label the image abc. how are the values in the ordered pairs affected by the translation? b translate △abc 5 units down. label the image abc. how are the values in the ordered pairs affected by the translation? c how could you determine the coordinates of the vertices of a translated image without using a graph?

Explanation:

Step1: Identify Original Coordinates

From the graph, find coordinates of \( A \), \( B \), \( C \). Let's assume \( A(-6,7) \), \( B(-4,2) \), \( C(-8,2) \) (from the table sketch).

Step2: Translate 3 Units Right (Part a)

For a horizontal translation \( h \) units right, new \( x \)-coordinate is \( x + h \), \( y \)-coordinate remains.

• \( A'(-6 + 3, 7) = (-3,7) \)? Wait, no, wait the table has \( A'(0,7) \)? Wait maybe original \( A(-6,7) \), right 3: \( -6 + 3 = -3 \)? Wait maybe my initial assumption is wrong. Wait the grid: let's check the graph. The left graph: \( A \) is at \( x=-6 \), \( y=7 \); \( B \) at \( x=-4 \), \( y=2 \); \( C \) at \( x=-8 \), \( y=2 \). Translating 3 right: add 3 to \( x \):

\( A'(-6 + 3, 7) = (-3,7) \)? But the table on the right has \( A'(0,7) \). Maybe original \( A(-6,7) \), right 6? No, the problem says 3 units right. Let's correct:
Translation rule: right 3: \( (x,y) \to (x + 3, y) \).
So \( A(-6,7) \to A'(-6 + 3,7) = (-3,7) \)
\( B(-4,2) \to B'(-4 + 3,2) = (-1,2) \)
\( C(-8,2) \to C'(-8 + 3,2) = (-5,2) \). Wait the table has different numbers, maybe the original coordinates are different. Wait the table has \( A(-6,7) \), \( A'(0,7) \): that's +6, so maybe the problem's table is a draft. Anyway, the key is: translating right 3 units affects the \( x \)-coordinate (increases by 3), \( y \)-coordinate stays.

Step3: Translate 5 Units Down (Part b)

Translation rule: down 5: \( (x,y) \to (x, y - 5) \).
So \( A(-6,7) \to A'(-6,7 - 5) = (-6,2) \)
\( B(-4,2) \to B'(-4,2 - 5) = (-4,-3) \)
\( C(-8,2) \to C'(-8,2 - 5) = (-8,-3) \). This affects the \( y \)-coordinate (decreases by 5), \( x \)-coordinate stays.

Step4: Determine Coordinates Without Graph (Part c)

Use translation rules: for horizontal translation \( h \) (right +, left -), vertical translation \( k \) (up +, down -), new coordinates are \( (x + h, y + k) \). So without graph, apply the translation vector (h,k) to each original vertex's coordinates.

Answer:

Part (a)

• Image \( A'B'C' \): Translate each vertex 3 units right. For \( A(x_1,y_1) \), \( B(x_2,y_2) \), \( C(x_3,y_3) \), new coordinates: \( A'(x_1 + 3, y_1) \), \( B'(x_2 + 3, y_2) \), \( C'(x_3 + 3, y_3) \).
• Effect: \( x \)-coordinates increase by 3, \( y \)-coordinates remain.

Part (b)

• Image \( A''B''C'' \): Translate each vertex 5 units down. New coordinates: \( A'(x_1, y_1 - 5) \), \( B'(x_2, y_2 - 5) \), \( C'(x_3, y_3 - 5) \).
• Effect: \( y \)-coordinates decrease by 5, \( x \)-coordinates remain.

Part (c)

Use the translation rule: if translating \( h \) units horizontally (right \( h>0 \), left \( h<0 \)) and \( k \) units vertically (up \( k>0 \), down \( k<0 \)), the new coordinates of a vertex \( (x,y) \) are \( (x + h, y + k) \). Apply this to each original vertex’s coordinates.

(Note: For specific coordinates, assume original \( A(-6,7) \), \( B(-4,2) \), \( C(-8,2) \):

• Part (a) \( A'(-3,7) \), \( B'(-1,2) \), \( C'(-5,2) \)
• Part (b) \( A''(-6,2) \), \( B''(-4,-3) \), \( C''(-8,-3) \))