Find the area of a parallelogram with the given vertices.

p(1, 3), q(3, 3), r(7, 8), s(9, 8)

Check the picture below

notice the length of the base, and of the height/altitude

Three ways to solve this problem:
1. Careful observation of the coordinates reveals that the side PQ is horizontal along the line y=3, and the side length is just (xq-xp)=(3-1)=2 (base =2)
Similarly, RS must also be horizontal with length 2 and along the line y=8.  (check)
Distance between lines RS and PQ = (yr-ys)=(8-3)=5 (height=5)
Area of parallelogram = base * height = 2*5=10

2. This method works for any convex figure with the vertices in order.
The vertices PQSR are in order, if we plot them on a graph.
Then we write the coordinates of PQSR vertically, and repeat P at the bottom, giving
1,3
3,3
9,8
7,8
1,3
Next, multiply diagonally down on each line and add together (4 times)
S1=1*3+3*8+9*8+7*3=3+24+72+21=120
similarly, multiply diagonally up, starting from the second line:
S2=3*3+9*3+7*8+1*8=9+27+56+8=100
The area is half of the bigger number less the smaller,
Area=(120-100)/2=10

3. By graphing on a square paper, and count squares.  Cannot do it here because there is no means to show the graph.  You are welcome use this method to check previous calculations.