Problem:
Two nonhorizontal, non vertical lines in the xy-coordinate plane intersect to form a 45∘ angle. One line has slope equal to 6 times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
Answer Choices:
A. 61
B. 32
C. 23
D. 3
E. 6 Solution:
Let the intersection point is the origin. Let (a,b) be a point on the line of lesser slope. The mutliplication of a+bi by cis 45. (a+bi)(21+i∗21)=21((a−b)+(a+b)∗i) and therefore (a−b,a+b) lies on the line of greater slope.
Then, the rotation of (a,b) by 45 degrees gives a line of slope a−ba+b.
We get the equation a6b=a−ba+b⟹a2−5ab+6b2=(a−3b)(a−2b)=0 and this gives our answer to be (C)23
OR
Intersect at the origin and select a point on each line to define vectors vi=(xi,yi). Note that θ=45∘ gives equal magnitudes of the vector products
v1⋅v2=v1v2cosθ and ∣v1×v2∣=v1v2sinθ
Substituting coordinate expressions for vector products, we find
v1⋅v2=∣v1×v2∣⟹x1x2+y1y2=x1y2−x2y1
Divide this equation by x1x2 to obtain
1+m1m2=m2−m1
where mi=yi/xi is the slope of line i. Taking m2=6m1, we obtain
6m12−5m1+1=0⟹m1∈{31, 21}
The latter solution gives the largest product of slopes m1m2=6m12=(C)23.