Question 305820
Find the # of square units contained in the region bordered by the graphs of the functions y=3x, x=3y, 3x+y=30?

put all 3 in y = mx + b form where m is the slope and b is the y-intercept (vertical intercept or point (0,b)):
1. y = 3x or y = 3x + 0
2. x = 3y --> x/3 = y or y = (1/3)x + 0
3. 3x + y = 30 --> y = -3x + 30

1/3 = -1 * 1/(-3) = -1 * -1/3 = 1/3 so equation 2 and 3 are perpendicular (the slopes are negative reciprocals)

{{{ graph( 300, 300, -5, 20, -5, 20, 3*x, x/3, -3*x + 30) }}}

this is a right triangle
equation 1 and 2 intersect at (0,0)
equation 2 and 3 intersect at:
(1/3)x = -3x + 30
x = -9x + 90
10x = 90
x = 9, eq.1 y = 3, eq. 2 y = 3 (9,3)
equation 1 and 3 intersect at:
3x = -3x + 30
6x = 30
x = 5, eq.1 y = 15, eq. 2 y = 15 (5,15)
from (0,0) to (9,3) is the base length is:
base = sqrt((9-0)^2 + (3-0)^2) = sqrt(9^2 + 3^2)
base = sqrt(81 + 9) = sqrt(90) = 3sqrt(10)
from (9,3) to (5,15) is the height length is:
height = sqrt((5-9)^2 + (15-3)^2) = sqrt((-4)^2 + 12^2)
height = sqrt(16 + 144) = sqrt(160) = 4sqrt(10)
area = 1/2 * base * height = 1/2 * 3sqrt(10) * 4sqrt(10)
area = 2 * 3 * 10 = 6 * 10 = 60 square units