Point to Line Homework Help (its a long question, btw)?

Question

Find the bisectors of the angles formed by the lines (0,-4); (5,-1) and (-1,-6); (0,1).

Here's what I did so far: I found the slope of the 2 lines. For Ln1 I got a slope of (3/5) and for Ln2 I got a slope of 7.  Then I wrote the equation for each line. Ln1 I has the equation y+4= (3/5)x. Ln2 I has the equation y-1=7x. I used point-slope form for both of these equations. Next, I set up the Point to Line formula. So I have: 

ab. value (3x-5y-20) divided by the sq. root of 3sq + 5sq.
EQUALS
ab. value (7x-y+1) divided by the sq. root of 7sq + 1sq.

Now this is where I'm confused. In the example that my teacher showed us, the lines were perpendicular and the bottom of the equations were equal, so they crossed out. But in this problem, the lines aren't perpendicular, and the bottoms of the equation work out to be RADICAL 34 and RADICAL 50.

So my question is (if you understand all of that) how do you do the problem from there? Or did I just mess the whole problem up? Please Help!

bippin · Accepted Answer

Here's how I solved it:
1. Determine the slope of the lines as you did
2. Determine the equations of the lines:  y=0.6x-4 and y=7x+1
3. Set these two formulas equal to each other in order to determine the point that they intersect at.   0.6x-4=7x+1
x=-25/32
(using one of the previous equations plug this x in to determine the coordinating y ) 
y=-143/32
4. Determine the angle between the two lines: (z is the angle)
tan(z)=(7-0.6)/(1+(0.6)(7))     
z=50.906 degrees
5. Divide the angle in half: z/2=25.453
6. tan(25.453)=(7-m)/(1+(m)(7))
m=1.506 = slope of the bisecting line
7. Solve y=mx+b using the new slope and the intersection point.    -143/32=-25/32(1.506)+b
b=-3.292
8. The bisecting line is:
        y=1.506x-3.292

This may not be the way that your teacher was teaching...but it works, and if you graph it you can verify the equations.

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Point to Line Homework Help (its a long question, btw)?

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