Problem:
Line ℓ1​ has equation 3x−2y=1 and goes through A=(−1,−2). Line ℓ2​ has equation y=1 and meets line ℓ1​ at point B. Line ℓ3​ has positive slope, goes through point A, and meets ℓ2​ at point C. The area of △ABC is 3 . What is the slope of ℓ3​ ?
Answer Choices:
A. 32​
B. 43​
C. 1
D. 34​
E. 23​
Solution:
The solution to the system of equations 3x−2y=1 and y=1 is B=(x,y)=(1,1). The perpendicular distance from A to BC is 3 . The area of △ABC is 21​⋅3⋅BC=3, so BC=2. Thus point C is 2 units to the right or to the left of B=(1,1). If C=(−1,1) then the line AC is vertical and the slope is undefined. If C=(3,1), then the line AC has slope 3−(−1)1−(−2)​=43​​.
The problems on this page are the property of the MAA's American Mathematics Competitions