Question

1. given the graph below, find pq.

Explanation:

To solve for the length of \( PQ \), we can use the distance formula or count the horizontal and vertical distances between points \( P \) and \( Q \) (assuming a grid where each square has side length 1). However, since the image is a bit unclear, we'll assume a typical grid-based approach.

Step 1: Determine the coordinates of \( P \) and \( Q \)

Let's assume the grid has integer coordinates. From the graph, let's estimate the coordinates:

• Let \( P = (-5, 2) \) (left side of the vertical line)
• Let \( Q = (1, -1) \) (right side of the vertical line)

Step 2: Calculate the horizontal and vertical differences

The horizontal difference (change in \( x \)): \( \Delta x = 1 - (-5) = 6 \)
The vertical difference (change in \( y \)): \( \Delta y = -1 - 2 = -3 \)

Step 3: Apply the distance formula

The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substituting the values:
\[
PQ = \sqrt{(6)^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \approx 6.71
\]

Wait, maybe the grid is simpler. Let's check again. If we count the horizontal squares between \( P \) and \( Q \): from \( x = -5 \) to \( x = 1 \), that's 6 units. Vertical: from \( y = 2 \) to \( y = -1 \), that's 3 units down. So using the Pythagorean theorem (since it's a right triangle):
\[
PQ = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \approx 6.71
\]

Alternatively, if the grid has each square as 1 unit, and the horizontal distance is 6, vertical is 3, then \( PQ = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5} \).

But maybe the coordinates are different. Let's re-examine the graph. If the vertical line is the y-axis (x=0), then:

• \( P \) is at \( (-4, 1) \)
• \( Q \) is at \( (1, -1) \)

Then \( \Delta x = 1 - (-4) = 5 \), \( \Delta y = -1 - 1 = -2 \). Then \( PQ = \sqrt{5^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29} \approx 5.39 \). But this is inconsistent.

Wait, maybe the horizontal distance is 5 and vertical is 3? Let's see. If \( P \) is at \( (-4, 2) \) and \( Q \) is at \( (1, -1) \), then \( \Delta x = 5 \), \( \Delta y = -3 \). Then \( PQ = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.83 \).

Alternatively, maybe the grid is such that the horizontal distance is 6 and vertical is 2? Then \( PQ = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \approx 6.32 \).

But since the original problem's graph is a bit unclear, let's assume the standard grid where each square is 1 unit, and the horizontal distance is 6, vertical is 3. Then \( PQ = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5} \approx 6.71 \).

Alternatively, if the horizontal distance is 5 and vertical is 3, then \( PQ = \sqrt{5^2 + 3^2} = \sqrt{34} \approx 5.83 \).

But maybe the correct coordinates are \( P(-5, 2) \) and \( Q(1, -1) \), so \( PQ = \sqrt{(1 - (-5))^2 + (-1 - 2)^2} = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \approx 6.71 \).

So the length of \( PQ \) is \( 3\sqrt{5} \) or approximately 6.71.

Final Answer

\( \boxed{3\sqrt{5}} \) (or approximately \( \boxed{6.71} \) if a decimal is preferred)

Answer:

To solve for the length of \( PQ \), we can use the distance formula or count the horizontal and vertical distances between points \( P \) and \( Q \) (assuming a grid where each square has side length 1). However, since the image is a bit unclear, we'll assume a typical grid-based approach.

Step 1: Determine the coordinates of \( P \) and \( Q \)

Let's assume the grid has integer coordinates. From the graph, let's estimate the coordinates:

• Let \( P = (-5, 2) \) (left side of the vertical line)
• Let \( Q = (1, -1) \) (right side of the vertical line)

Step 2: Calculate the horizontal and vertical differences

The horizontal difference (change in \( x \)): \( \Delta x = 1 - (-5) = 6 \)
The vertical difference (change in \( y \)): \( \Delta y = -1 - 2 = -3 \)

Step 3: Apply the distance formula

The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
Substituting the values:
\[
PQ = \sqrt{(6)^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \approx 6.71
\]

Wait, maybe the grid is simpler. Let's check again. If we count the horizontal squares between \( P \) and \( Q \): from \( x = -5 \) to \( x = 1 \), that's 6 units. Vertical: from \( y = 2 \) to \( y = -1 \), that's 3 units down. So using the Pythagorean theorem (since it's a right triangle):
\[
PQ = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \approx 6.71
\]

Alternatively, if the grid has each square as 1 unit, and the horizontal distance is 6, vertical is 3, then \( PQ = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5} \).

But maybe the coordinates are different. Let's re-examine the graph. If the vertical line is the y-axis (x=0), then:

• \( P \) is at \( (-4, 1) \)
• \( Q \) is at \( (1, -1) \)

Then \( \Delta x = 1 - (-4) = 5 \), \( \Delta y = -1 - 1 = -2 \). Then \( PQ = \sqrt{5^2 + (-2)^2} = \sqrt{25 + 4} = \sqrt{29} \approx 5.39 \). But this is inconsistent.

Wait, maybe the horizontal distance is 5 and vertical is 3? Let's see. If \( P \) is at \( (-4, 2) \) and \( Q \) is at \( (1, -1) \), then \( \Delta x = 5 \), \( \Delta y = -3 \). Then \( PQ = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.83 \).

Alternatively, maybe the grid is such that the horizontal distance is 6 and vertical is 2? Then \( PQ = \sqrt{6^2 + 2^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \approx 6.32 \).

But since the original problem's graph is a bit unclear, let's assume the standard grid where each square is 1 unit, and the horizontal distance is 6, vertical is 3. Then \( PQ = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5} \approx 6.71 \).

Alternatively, if the horizontal distance is 5 and vertical is 3, then \( PQ = \sqrt{5^2 + 3^2} = \sqrt{34} \approx 5.83 \).

But maybe the correct coordinates are \( P(-5, 2) \) and \( Q(1, -1) \), so \( PQ = \sqrt{(1 - (-5))^2 + (-1 - 2)^2} = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \approx 6.71 \).

So the length of \( PQ \) is \( 3\sqrt{5} \) or approximately 6.71.

Final Answer

\( \boxed{3\sqrt{5}} \) (or approximately \( \boxed{6.71} \) if a decimal is preferred)