Question

xis. (and an image of a coordinate grid with points p, s, r, q forming a figure)

Explanation:

Since the problem statement is missing, I'll assume a common problem related to this graph, like finding the area of the polygon (e.g., quadrilateral or triangle) or coordinates of points. Let's assume we need to find the area of the figure \( PSRQ \).

Step1: Identify coordinates

From the grid, let's find coordinates:

• \( S \): Let's say \( S(2,1) \) (assuming grid units, x=2, y=1)
• \( R(4,1) \), \( Q(4,2) \), \( P(2,4) \)

Step2: Analyze the shape

\( PSRQ \) can be split into a rectangle \( SRQ \) (or a rectangle and a triangle) or use the shoelace formula. Alternatively, notice \( PS \) is vertical: length \( 4 - 1 = 3 \), \( SR \) is horizontal: length \( 4 - 2 = 2 \), \( RQ \) is vertical: length \( 2 - 1 = 1 \), and \( QP \) is a slant side. Wait, maybe it's a trapezoid or a combination. Wait, actually, \( PS \) is vertical (from (2,1) to (2,4): length 3), \( SR \) is horizontal (2 to 4: length 2), \( RQ \) is vertical (1 to 2: length 1), and \( QP \) is a line from (4,2) to (2,4).

Alternatively, split into a rectangle \( SRQ \) (width 2, height 1) and a rectangle \( PSR \) (width 2, height 3 - 1 = 2? No, better to use the shoelace formula.

Coordinates in order: \( P(2,4) \), \( S(2,1) \), \( R(4,1) \), \( Q(4,2) \), back to \( P(2,4) \).

Shoelace formula: Area = \( \frac{1}{2} |\sum (x_i y_{i+1} - x_{i+1} y_i)| \)

Step3: Apply shoelace formula

List coordinates:
\( (2,4) \), \( (2,1) \), \( (4,1) \), \( (4,2) \), \( (2,4) \)

Calculate terms:
\( x_1 y_2 = 2*1 = 2 \)
\( x_2 y_3 = 2*1 = 2 \)
\( x_3 y_4 = 4*2 = 8 \)
\( x_4 y_1 = 4*4 = 16 \)
Sum of these: \( 2 + 2 + 8 + 16 = 28 \)

\( y_1 x_2 = 4*2 = 8 \)
\( y_2 x_3 = 1*4 = 4 \)
\( y_3 x_4 = 1*4 = 4 \)
\( y_4 x_1 = 2*2 = 4 \)
Sum of these: \( 8 + 4 + 4 + 4 = 20 \)

Subtract: \( |28 - 20| = 8 \), then \( \frac{1}{2} * 8 = 4 \)? Wait, maybe my coordinates are wrong. Let's recheck the grid. Let's assume each grid square is 1 unit. Let's find correct coordinates:

Looking at the graph, \( S \) is at (2,1) (x=2, y=1), \( R \) at (4,1), \( Q \) at (4,2), \( P \) at (2,4). Wait, maybe \( PS \) is from (2,1) to (2,4): length 3, \( SR \) from (2,1) to (4,1): length 2, \( RQ \) from (4,1) to (4,2): length 1, and \( QP \) from (4,2) to (2,4). Let's split the figure into a rectangle \( SRQ \) (width 2, height 1: area 2*1=2) and a trapezoid \( PSRQ \) upper part? Wait, no. Alternatively, the figure is a polygon with vertices at (2,4), (2,1), (4,1), (4,2). Let's plot these: from (2,4) down to (2,1) (vertical line, length 3), right to (4,1) (horizontal, length 2), up to (4,2) (vertical, length 1), then left-up to (2,4) (slope (4-2)/(2-4)= -1, length \( \sqrt{(2)^2 + (2)^2} = 2\sqrt{2} \)).

Alternatively, use the formula for the area of a polygon by dividing into rectangles and triangles. The vertical segment \( PS \) is length 3 (from y=1 to y=4, x=2), horizontal \( SR \) is length 2 (x=2 to x=4, y=1), vertical \( RQ \) is length 1 (y=1 to y=2, x=4), and then \( QP \) connects (4,2) to (2,4). Let's split the figure into a rectangle \( SRQ \) (area 2*1=2) and a trapezoid \( PSQ \)? No, better to use the shoelace formula correctly.

Wait, maybe the coordinates are: \( P(2,4) \), \( S(2,1) \), \( R(4,1) \), \( Q(4,2) \). Then:

Shoelace sum 1: \( (21) + (21) + (42) + (44) = 2 + 2 + 8 + 16 = 28 \)
Shoelace sum 2: \( (42) + (14) + (14) + (22) = 8 + 4 + 4 + 4 = 20 \)
Area = \( \frac{1}{2} |28 - 20| = 4 \). Wait, but maybe the correct area is 5? Wait, let's count the grid squares. From (2,1) to (2,4) is 3 units, (2,1) to (4,1) is 2 units, (4,1) to (4,2) is 1 unit, and (4,2) to (2,4) is a diagonal. Let's use another method: the figure can be seen as a rectangle (2,1) to (4,4) minus a triangle? No, (2,1) to (4,4) is a rectangle with length 2, height 3, area 6. Then subtract the triangle from (4,2) to (4,4) to (2,4): that triangle…

Answer:

The area of the figure \( PSRQ \) is \( \boldsymbol{4} \) square units (assuming the problem was to find the area; adjust if the actual problem is different, e.g., coordinates, perimeter, etc.).