Problem:
Line β1β has equation 3xβ2y=1 and goes through A=(β1,β2). Line β2β has equation y=1 and meets line β1β at point B. Line β3β has positive slope, goes through point A, and meets β2β at point C. The area of β³ABC is 3 . What is the slope of β3β ?
Answer Choices:
A. 32β
B. 43β
C. 1
D. 34β
E. 23β
Solution:
The solution to the system of equations 3xβ2y=1 and y=1 is B=(x,y)=(1,1). The perpendicular distance from A to BC is 3 . The area of β³ABC is 21ββ
3β
BC=3, so BC=2. Thus point C is 2 units to the right or to the left of B=(1,1). If C=(β1,1) then the line AC is vertical and the slope is undefined. If C=(3,1), then the line AC has slope 3β(β1)1β(β2)β=43ββ.
The problems on this page are the property of the MAA's American Mathematics Competitions