Find an equation of the line that bisects the obtuse angles formed by the lines with equations 3x-y=1 and x+y=-2.
a. (3√2 +√10)x-(√10 + √2)y-2√10+√2=0
b. (3√2 - √10)x+(√10 - √2)y+2√10+√2=0
c. (3√2 + √10)x+(√10 - √2)y+2√10 - √2=0
d. (3√2 + √10)x-(√10 + √2)y-2√10-√2=0
i'm not sure on this one!
before I answer this question, I have to establish if you know
1. that slope = tan(angle the line makes with the x-axis)
2. if m1 and m2 are the slopes of 2 nonparallel lines, and theta is the acute angle between them, then
tan(theta) = │(m1 - m2)/(1 + m1m2)│
If not please watch this short video.
3. tan(2A) = 2tanA/(1-tan^2 A)
I also noticed that in the first answer given, the slope of the line is
(3√2+√10)/(√10+√2) which when rationalized is (1+√5)/2 which is the answer that I got for the required line.
So you might have obtained the correct slope of the new line without realizing it.
Let me know if you need more help.posted by Reiny
okay thanksposted by sophie