Problem:
The equations of L1β and L2β are y=mx and y=nx, respectively. Suppose L1β makes twice as large an angle with the horizontal (measured counterclockwise from the positive x-axis) as does L2β, and that L1β has 4 times the slope of L2β. If L1β is not horizontal, then mn is
Answer Choices:
A. 22ββ
B. β22ββ
C. 2
D. β2
E. not uniquely determined by the given information
Solution:
In the adjoining figure, L1β and L2β 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 L2β and BC is the slope of L1β. Therefore, AC=n,BC=m, and AB=3n. Since OA is an angle bisector
OBOCβ=ABACβ
This yields
OB1β=3nnβ and OB=3
By the Pythagorean theorem 1+(4n)2=9, so n=22ββ. Since m=4n, mn=4n2=2.
OR
Let ΞΈ1β and ΞΈ2β be the angles of inclination of lines L1β and L2β, respectively. Then m=tanΞΈ1β and n=tanΞΈ2β. Since ΞΈ1β=2ΞΈ2β and m=4n,4n=m=tanΞΈ1β=tan2ΞΈ2β=1βtan2ΞΈ2β2tanΞΈ2ββ=1βn22nβ.
Thus 4n(1βn2)=2n. Since nξ =0,2n2=1, and mn, which equals 4n2, is 2.
