What is the slope of the bisector of angle ABC?

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To find the slope of the bisector of angle ABC formed by points A (1,2), B (0,0), and C (-1,3), the lengths and slopes of segments BA, BC, and AC have been calculated. The slopes of BA and BC are 2 and -3, respectively, and segment AC has a slope of -1/2, indicating that triangle CAB is a right triangle. The tangent of angle ABC can be determined using the lengths of the segments, leading to a tangent value of √3. Utilizing this information and trigonometric identities will help in deriving the slope of the angle bisector.
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Homework Statement



A, B, and C, are points (1,2), (0,0), and (-1,3) respectively.

Homework Equations



Find the slope of the bisector of angle ABC.

The Attempt at a Solution



I have done the following and don't know how it helpsL

-found length of BA, BC, and AC
-found the slope of BA, and BC
-i don't know where to go from here

The length of BA is √5

The length of BC is √10

The length of AC is √5

The slope of BA is 2

The slope of BC is -3

Please do not do the work for me, i just need some advice as to what to do from here.
 
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P.S. this is probably not precalculus, but this is problem that was in my precalculus book (i think we a re reviewing from algebra).

i forgot so much from last year:confused:
 
I'm not quite sure how'd one would go about doing this problem without a bit of trigonometry, but for starters, the segment AC has slope -1/2 and is therefore perpendicular to AB. Hence CAB is a right triangle (i.e. angle CAB is a right angle). This forces the distance of AC to be sqrt(15) (This is verifiable even if CAB isn't right).

But since CAB is a right triangle then the tangent of angle ABC is \sqrt(15)/\sqrt(5) =\sqrt(3). This along with a well timed tangent angle sum indentity should get you the answer.

--Elucidus
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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