Problem:
Two nonhorizontal, non vertical lines in the xy-coordinate plane intersect to form a 45β 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. 61β
B. 32β
C. 23β
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)(2β1β+iβ2β1β)=2β1β((aβb)+(a+b)βi) and therefore (aβb,a+b) lies on the line of greater slope.
Then, the rotation of (a,b) by 45 degrees gives a line of slope aβba+bβ.
We get the equation a6bβ=aβba+bββΉa2β5ab+6b2=(aβ3b)(aβ2b)=0 and this gives our answer to be (C)23ββ
OR
Intersect at the origin and select a point on each line to define vectors viβ=(xiβ,yiβ). Note that ΞΈ=45β gives equal magnitudes of the vector products
v1ββ v2β=v1βv2βcosΞΈ and β£v1βΓv2ββ£=v1βv2βsinΞΈ
Substituting coordinate expressions for vector products, we find