Expert: Sherry Wallin - 11/30/2008

Question
Find the angle (* theta) between two non-vertical lines L1 and L2. The angle
(*theta) satisfies the equation

tan (*theta) = |m2-m1/1 m2m1|

where m1 and m2 are the slopes of L1 and L2, respectively. (Assume that
m1m2 does not equal 1)

L1: 3x-2y=5
L2: x y=1


I've tried and tried but i keep coming up with too short of an answer. How
would you do this type of problem

Answer
Hi Cheyenne~
    I am going to assume that the tan(theta) = |(m2-m1)/m2*m1|
You say to assume m1m2 does not equal 1, this is not clear, I think you mean m1m2 is not equal to 0 because we do not divide by 0 since  it is undefined. So assuming you mean m1m2 is not equal 0 we proceed as follows:

Let the slope m1 be the slope of L1, simply solve for y and use the coefficient of x like so: 3x - 2y = 5
                         -3x       -3x
                              -2y = -3x + 5
                         (-2/-2)y = (-3/-2)x + 5/-2
                                y = (3/2)x - 5/2
m1 = 3/2
likewise for L2
Here again there is a plus sign missing from the equation I believe
I think the equation is x "plus" y = 1 so y = -x + 1 giving us
m2 = -1

|[(3/2)-(-1)]/(3/2)(-1)| = |(5/2)/(-3/2)|= |5/-3| = 5/3
so tan (theta) = 5/3, take the tan^-1 (theta) both sides of the equation and you are finding tan^-1(5/3) = 59 degrees and a little bit more.
Bear in mind that there are two possible answers for this and remember that the tan (theta) = sin(theta)/cos(theta)and so you want the quadrant where both x and y have the same sign so that you have a positive value for the tangent function. So 59 degrees is in the first quadrant and you want the angle in the 3rd quadrant that has tan x degrees = 5/3 which is 180 degrees + 59 degrees or 239 degrees.

Best Wishes for your studies,
Math Prof