Question

prove that the diagonals of kite uvwx are perpendicular.
step 1: determine the slope of $overline{xv}$.
the slope of $overline{xv}$ is $square$.
step 2: determine the slope of $overline{uw}$.
the slope of $overline{uw}$ is $square$.
step 3: the slopes of the diagonals are $square$.
the diagonals of kite uvwx are $square$.

Explanation:

To solve this, we first identify the coordinates of the vertices from the graph. Let's assume the coordinates (by analyzing the grid):

• \( X(-1, -2) \), \( V(5, 4) \), \( U(-3, 5) \), \( W(7, -5) \)

Step 1: Slope of \( \overline{XV} \)

The slope formula is \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
For \( X(-1, -2) \) and \( V(5, 4) \):
\( m_{XV} = \frac{4 - (-2)}{5 - (-1)} = \frac{6}{6} = 1 \)

Step 2: Slope of \( \overline{UW} \)

For \( U(-3, 5) \) and \( W(7, -5) \):
\( m_{UW} = \frac{-5 - 5}{7 - (-3)} = \frac{-10}{10} = -1 \)

Step 3: Relationship of Slopes

Two lines are perpendicular if the product of their slopes is \( -1 \).
\( m_{XV} \times m_{UW} = 1 \times (-1) = -1 \), so the slopes are negative reciprocals (or their product is \( -1 \)).

Conclusion

Since the product of the slopes of the diagonals is \( -1 \), the diagonals of kite \( UVWX \) are perpendicular.

Filling in the Blanks:

• Slope of \( \overline{XV} \): \( \boldsymbol{1} \)
• Slope of \( \overline{UW} \): \( \boldsymbol{-1} \)
• The slopes of the diagonals are \( \boldsymbol{\text{negative reciprocals (or product is } -1)} \)
• The diagonals of kite \( UVWX \) are \( \boldsymbol{\text{perpendicular}} \)

Answer:

To solve this, we first identify the coordinates of the vertices from the graph. Let's assume the coordinates (by analyzing the grid):

• \( X(-1, -2) \), \( V(5, 4) \), \( U(-3, 5) \), \( W(7, -5) \)

Step 1: Slope of \( \overline{XV} \)

The slope formula is \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
For \( X(-1, -2) \) and \( V(5, 4) \):
\( m_{XV} = \frac{4 - (-2)}{5 - (-1)} = \frac{6}{6} = 1 \)

Step 2: Slope of \( \overline{UW} \)

For \( U(-3, 5) \) and \( W(7, -5) \):
\( m_{UW} = \frac{-5 - 5}{7 - (-3)} = \frac{-10}{10} = -1 \)

Step 3: Relationship of Slopes

Two lines are perpendicular if the product of their slopes is \( -1 \).
\( m_{XV} \times m_{UW} = 1 \times (-1) = -1 \), so the slopes are negative reciprocals (or their product is \( -1 \)).

Conclusion

Since the product of the slopes of the diagonals is \( -1 \), the diagonals of kite \( UVWX \) are perpendicular.

Filling in the Blanks:

• Slope of \( \overline{XV} \): \( \boldsymbol{1} \)
• Slope of \( \overline{UW} \): \( \boldsymbol{-1} \)
• The slopes of the diagonals are \( \boldsymbol{\text{negative reciprocals (or product is } -1)} \)
• The diagonals of kite \( UVWX \) are \( \boldsymbol{\text{perpendicular}} \)