SOLUTION: Find the equation of the bisector of the acute angles and also the equation of the bisector of the obtuse angles formed by the line x+y =2 and x + 2y = 3.

Algebra ->  Testmodule -> SOLUTION: Find the equation of the bisector of the acute angles and also the equation of the bisector of the obtuse angles formed by the line x+y =2 and x + 2y = 3.      Log On


   

Question 374060: Find the equation of the bisector of the acute angles and also the equation of the bisector of the obtuse angles formed by the line x+y =2 and x + 2y = 3.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Let (x,y) be a point in Quadrant 4 that is located in the angle bisector of the acute angle between the two lines. We use the distance formula between a point and a line: d = |Ax + By + C|/sqrt%28A%5E2+%2B+B%5E2%29.
Then the point (x,y)is BELOW the line x + 2y = 3, so its distance from the said line is d+=+-%28x%2B2y-3%29%2Fsqrt%281%5E2+%2B+2%5E2%29+=+-%28x%2B2y-3%29%2Fsqrt%285%29.
The point (x,y)is ABOVE the line x + y = 2, so its distance from the said line is d+=+%28x%2By-2%29%2Fsqrt%281%5E2+%2B+1%5E2%29+=+%28x%2By-2%29%2Fsqrt%282%29.
Hence -%28x%2B2y-3%29%2Fsqrt%285%29+=+%28x%2By-2%29%2Fsqrt%282%29;
. This the equation of the angle bisector of the ACUTE angle of intersection.
The arguments for the equation of the angle bisector of the OBTUSE angle of intersection run along the same lines. (But choose (x,y) now to be in Q1, and such point will be ABOVE both lines, so use +'s in place of the absolute value bars.)