Find an equation of the line that bisects the obtuse angles formed by the lines with equations 3x-y=1 and x+y=-2.

Question

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!

Answer 1

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.

http://www.slideshare.net/nsimmons/11-x1-t05-07-angle-between-two-lines

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.

Answer 2

okay thanks