Problem:
The equations of L1 and L2 are y=mx and y=nx, respectively. Suppose L1 makes twice as large an angle with the horizontal (measured counterclockwise from the positive x-axis) as does L2, and that L1 has 4 times the slope of L2. If L1 is not horizontal, then mn is
Answer Choices:
A. 22
B. −22
C. 2
D. −2
E. not uniquely determined by the given information
Solution:
In the adjoining figure, L1 and L2 intersect the line x=1 at B and A, respectively; C is the intersection of the line x=1 with the x-axis. Since OC=1, AC is the slope of L2 and BC is the slope of L1. Therefore, AC=n,BC=m, and AB=3n. Since OA is an angle bisector
OBOC=ABAC
This yields
OB1=3nn and OB=3
By the Pythagorean theorem 1+(4n)2=9, so n=22. Since m=4n, mn=4n2=2.
OR
Let θ1 and θ2 be the angles of inclination of lines L1 and L2, respectively. Then m=tanθ1 and n=tanθ2. Since θ1=2θ2 and m=4n,4n=m=tanθ1=tan2θ2=1−tan2θ22tanθ2=1−n22n.
Thus 4n(1−n2)=2n. Since n=0,2n2=1, and mn, which equals 4n2, is 2.
