Problem:
Two lines with slopes 21β and 2 intersect at (2,2). What is the area of the triangle enclosed by these two lines and the line x+y=10?
Answer Choices:
A. 4
B. 42β
C. 6
D. 8
E. 62β Solution:
Let's first work out the slope-intercept form of all three lines: (x,y)=(2,2) and y=2xβ+b implies 2=22β+b=1+b so b=1, while y=2x+c implies 2=2β 2+c=4+c so c=β2. Also, x+y=10 implies y=βx+10. Thus the lines are y=2xβ+1,y=2xβ2, and y=βx+10. Now we find the intersection points between each of the lines with y=βx+10, which are (6,4) and (4,6). Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base 22β and height 32β, whose area is (C)6β.
OR
Like in other solutions, we find that the three points of intersection are (2,2),(4,6), and (6,4). By the Pythagorean theorem, this is an isosceles triangle with base 22β and equal length 25β. The area of an isosceles triangle with base b and equal length l is 4b4l2βb2ββ. Plugging in b=22β and l=25β,