1. The lines y=x/3 and y=13x/9 are drawn in the coordinate plane. Find the slope of the line that bisects the angle between these lines.
2. The line y=(3x+15)/4 intersects circle x^2+y^2=36 at points Q and R. Find the length of QR.
For question number one, right triangles seem useful because slope is "rise over run", which, if you graph, creates a right triangle. But, otherwise, I'm very confused.
+36916 Y= mx + b is the form of a line where m = slope
y = 1/3 x +0 is of this form slope = 1/3
y= 13/9 x + 0 slope = 13/9
we need the angle between these two [arctan (1/3) + arctan (13/9)]/2 = 36.8699o
tan ( 36.8688) = slope of line bisecting other two = .75 = 3/4
so y = 3/4 x will bisect the angle of the other two ~ 
Here is a verification graph:
https://www.desmos.com/calculator/7gf4hgooic
+23246 2) x2 + y2 = 36 and y = (3x + 15)/4
x2 + [ (3x + 15) / 4 ]2 = 36
x2 + (9x2 + 90x + 225) / 16 = 36
16x2 + (9x2 + 90x + 225) = 576
25x2 + 90x - 351 = 0
Quadratic formula: x1 = [ -90 + sqrt[( 43 200 ) ] / 50 or x2 = [ -90 - sqrt[( 43 200 ) ] / 50
y1 = [ 480 + 3·sqrt( 43 200 ) ] / 200 y2 = [ 480 - 3·sqrt( 43 200 ) ] / 200
Distance formula: sqrt[ ( x2 - x1 )2 + ( y2 - y1 )2 ]
= 10.392 (calculator time)
