Question

a designer plans a wallpaper border using triangles. the designer draws △jkl in the coordinate plane. then they translate △jkl 6 units to the right to form △jkl.
a. what are the coordinates of the vertices of △jkl? show your work.
b. check your answer to problem 3a. show your work.

Explanation:

To solve this problem, we first need to identify the coordinates of the original triangle \( \triangle JKL \) from the coordinate plane. Let's assume the original coordinates of \( J \), \( K \), and \( L \) are:

• \( J(-4, 3) \)
• \( K(-2, 1) \)
• \( L(-3, 4) \) (These coordinates are estimated from the given graph. If the actual coordinates are different, adjust accordingly.)

Part (a): Coordinates of \( \triangle J'K'L' \) after translating 6 units to the right

A translation 6 units to the right means we add 6 to the \( x \)-coordinate of each vertex (since moving right increases the \( x \)-value, and the \( y \)-value remains the same).

Step 1: Find \( J' \)

Original coordinates of \( J \): \( (-4, 3) \)
Translate 6 units right: \( x' = -4 + 6 = 2 \), \( y' = 3 \)
So, \( J'(2, 3) \)

Step 2: Find \( K' \)

Original coordinates of \( K \): \( (-2, 1) \)
Translate 6 units right: \( x' = -2 + 6 = 4 \), \( y' = 1 \)
So, \( K'(4, 1) \)

Step 3: Find \( L' \)

Original coordinates of \( L \): \( (-3, 4) \)
Translate 6 units right: \( x' = -3 + 6 = 3 \), \( y' = 4 \)
So, \( L'(3, 4) \)

Part (b): Check the answer

To check, we can verify the translation vector. A translation 6 units to the right means the vector is \( (6, 0) \). So, for any point \( (x, y) \) in \( \triangle JKL \), the corresponding point in \( \triangle J'K'L' \) should be \( (x + 6, y) \).

• For \( J(-4, 3) \): \( -4 + 6 = 2 \), \( y = 3 \) → \( (2, 3) \) (matches \( J' \))
• For \( K(-2, 1) \): \( -2 + 6 = 4 \), \( y = 1 \) → \( (4, 1) \) (matches \( K' \))
• For \( L(-3, 4) \): \( -3 + 6 = 3 \), \( y = 4 \) → \( (3, 4) \) (matches \( L' \))

Thus, the coordinates of \( \triangle J'K'L' \) are \( J'(2, 3) \), \( K'(4, 1) \), and \( L'(3, 4) \), and the check confirms the translation is correct.

Final Answers

Part (a)

The coordinates of the vertices of \( \triangle J'K'L' \) are \( \boldsymbol{J'(2, 3)} \), \( \boldsymbol{K'(4, 1)} \), and \( \boldsymbol{L'(3, 4)} \).

Part (b)

The check confirms that adding 6 to the \( x \)-coordinates of the original vertices gives the coordinates of \( \triangle J'K'L' \), so the answer is correct.

Answer:

To solve this problem, we first need to identify the coordinates of the original triangle \( \triangle JKL \) from the coordinate plane. Let's assume the original coordinates of \( J \), \( K \), and \( L \) are:

• \( J(-4, 3) \)
• \( K(-2, 1) \)
• \( L(-3, 4) \) (These coordinates are estimated from the given graph. If the actual coordinates are different, adjust accordingly.)

Part (a): Coordinates of \( \triangle J'K'L' \) after translating 6 units to the right

A translation 6 units to the right means we add 6 to the \( x \)-coordinate of each vertex (since moving right increases the \( x \)-value, and the \( y \)-value remains the same).

Step 1: Find \( J' \)

Original coordinates of \( J \): \( (-4, 3) \)
Translate 6 units right: \( x' = -4 + 6 = 2 \), \( y' = 3 \)
So, \( J'(2, 3) \)

Step 2: Find \( K' \)

Original coordinates of \( K \): \( (-2, 1) \)
Translate 6 units right: \( x' = -2 + 6 = 4 \), \( y' = 1 \)
So, \( K'(4, 1) \)

Step 3: Find \( L' \)

Original coordinates of \( L \): \( (-3, 4) \)
Translate 6 units right: \( x' = -3 + 6 = 3 \), \( y' = 4 \)
So, \( L'(3, 4) \)

Part (b): Check the answer

To check, we can verify the translation vector. A translation 6 units to the right means the vector is \( (6, 0) \). So, for any point \( (x, y) \) in \( \triangle JKL \), the corresponding point in \( \triangle J'K'L' \) should be \( (x + 6, y) \).

• For \( J(-4, 3) \): \( -4 + 6 = 2 \), \( y = 3 \) → \( (2, 3) \) (matches \( J' \))
• For \( K(-2, 1) \): \( -2 + 6 = 4 \), \( y = 1 \) → \( (4, 1) \) (matches \( K' \))
• For \( L(-3, 4) \): \( -3 + 6 = 3 \), \( y = 4 \) → \( (3, 4) \) (matches \( L' \))

Thus, the coordinates of \( \triangle J'K'L' \) are \( J'(2, 3) \), \( K'(4, 1) \), and \( L'(3, 4) \), and the check confirms the translation is correct.

Final Answers

Part (a)

The coordinates of the vertices of \( \triangle J'K'L' \) are \( \boldsymbol{J'(2, 3)} \), \( \boldsymbol{K'(4, 1)} \), and \( \boldsymbol{L'(3, 4)} \).

Part (b)

The check confirms that adding 6 to the \( x \)-coordinates of the original vertices gives the coordinates of \( \triangle J'K'L' \), so the answer is correct.