+162 Two black lines pass through the origin, as shown below. If their slopes are 2 and 8, then find the slope of the red line that bisects the acute angle between these lines.
+130060 
Construct a circle centered at the origin with a radius of 1
Let the line with the slope of 8 be y =8x = (8/1)x
Let the line with the slope of 2 be y = 2x = (2/1)x
Let the point where the line with the slope of 8x intersects the circle be
(1/sqrt (1^2 + 8^2) / 8/(sqrt (1^2 + 8^2)) =
( 1/sqrt65 , 8/sqrt 65 ) = A
Let the point where the line with the slope of 2x intersects the circle be
(1/sqrt (1^2 + 5^2) , 2/sqrt (1^2 + 2^2)) =
(1/sqrt 5 , 2/sqrt 5) = B
Draw AB
The midpoint of this segment is
[ (1/sqrt 65 + 1/sqrt 5) /2 , (8/sqrt 65 + 2/sqrt 5) /2 ] =
[ ( sqrt 5 + sqrt 65) /2 ,(8sqrt 5 + 2 sqrt 65) /2 ] =
[ (sqrt 5/ 2) ( 1 + sqrt 13) , (sqrt 5/ 2) ( 8 + 2sqrt 13) ] = C
The slope of the line through the origin through C is
[ 8 + 2sqrt 13 ] / [ (1 + sqrt 13 ] =
[ 8 + 2sqrt 13 ] [1 - sqrt 13 ] [ 1 -13] =
[ 8 + 2sqrt 13 - 8sqrt 13 - 26 ] /[ -12] =
[ -18 - 6sqrt13 ] / [-12] =
[ 3 + sqrt 13 ] / 2 =slope of line bisecting the acute angle between these lines
