Parallelogram has vertices ABCD, A(3,3), B(-3,-3), C(-4,3) and D(2,9). If a point is selected at random from the region determined by the parallelogram, what is the probability that the point is not above the x-axis? Express your answer as a common fraction.

Guest Dec 7, 2020

#1**0 **

First let us find the equation for the lines of AB and CD (the lines that cross the x axis). Then, we can find the area of the region under the x axis and put that over the area of total area. Probability is success/total, and in this case, the area under the x axis is "success", and the total area is the "total."

Line AB has points (3,3) and (-3,-3). Writing this in slope-intercept form, which is y=mx+b, we get:

3=3m+b

-3=-3m+b

Adding these equations we get:

0=2b, b=0.

Substituting b=0 into our original equations we get:

3=3m+0, m=1.

Therefore, the line for AB is:

y=x.

Now it is seemingly obvious that when the line crosses the x axis (when y is 0), it passes through point (0,0).

We can repeat the same steps we did for the first line for the second, and we will find that the equation of the second line (CD) is:

y=2x+5

Therefore, this line crosses the x axis when y is 0 and x is -5/2.

From this information, we can now find the area of the region under the x axis. This is a funky shape, so it would be easiest to draw a picture, find the area of a big rectangle around the figure, and subtract triangles until you get what is left. I found that the area of the region was 6.75, and the are of the big figure was 28.5.

Therefore the probability is 6.75/28.5 = 9/38. <-- My Answer

Guest Dec 7, 2020

#2**+1 **

A= (3,3) = (x1, y 1)

D = (2,9) = (x2, y2)

C = (-4,3) = (x3, y3)

B = (-3,-3)= (x4, y4)

Area = (1/2) [ ( x1y2 + x2y3 + x3y4 + x4y1) - (x2y1 + x3y2 + x4y3 + x1y4 ) ] =

(1/2) [ (27 + 6 + 12 -9) - ( 6 -36 - 9 -9) ] =

(1/2) [ 36 + 48 ] = 84/ 2 = 42

Area of triangle EFB =(1/2) (3.5) (3) = 5.25

P ( below the x axis ) = 5.25 / 42 = ( 21/4) / 42 = 1/8

CPhill Dec 7, 2020