Posted by **Sophie** on Tuesday, June 2, 2009 at 3:48pm.

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!

- Math -
**Reiny**, Tuesday, June 2, 2009 at 4:57pm
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.

- Math -
**sophie**, Tuesday, June 2, 2009 at 6:11pm
okay thanks

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