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Analytical Geometry
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Ambareesh Rishi
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Joined: 4 May 2007
Posts: 499
9 Apr 2008 14:44:40 IST
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answer is 2
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9 Apr 2008 17:29:34 IST
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let a1x+b1y+c1=0 and a2x+b2y+c2=0 are the lines.
then the procedure to find the angle bisectors is
1) make c2&c2 positive.
2)if a1a2+b1b2>0
=> (a1x+b1y+c1)/root(a12+a22) =+ (a2x+b2y+c2)/root(a22+b22)
is obtuse angle bisector
=>(a1x+b1y+c1)/root(a12+a22) =- (a2x+b2y+c2)/root(a22+b22)
is acute angle bisector
-----------------------------------------
if a1a2+b1b2<0
=> (a1x+b1y+c1)/root(a12+a22) =+ (a2x+b2y+c2)/root(a22+b22)
is acute angle bisector
=>(a1x+b1y+c1)/root(a12+a22) =- (a2x+b2y+c2)/root(a22+b22)
is obtuse angle bisector
then the procedure to find the angle bisectors is
1) make c2&c2 positive.
2)if a1a2+b1b2>0
=> (a1x+b1y+c1)/root(a12+a22) =+ (a2x+b2y+c2)/root(a22+b22)
is obtuse angle bisector
=>(a1x+b1y+c1)/root(a12+a22) =- (a2x+b2y+c2)/root(a22+b22)
is acute angle bisector
-----------------------------------------
if a1a2+b1b2<0
=> (a1x+b1y+c1)/root(a12+a22) =+ (a2x+b2y+c2)/root(a22+b22)
is acute angle bisector
=>(a1x+b1y+c1)/root(a12+a22) =- (a2x+b2y+c2)/root(a22+b22)
is obtuse angle bisector
9 Apr 2008 17:48:33 IST
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Some additional info:
Equations of the bisectors of the angles between two lines represented by ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 are given by
(x-x1)^2 - (y-y1)^2/a-b = (x-x1)(y-y1)/h
where(x1,y1) is the point of intersection of the lines represented by the given equation.
Equations of the bisectors of the angles between two lines represented by ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 are given by
(x-x1)^2 - (y-y1)^2/a-b = (x-x1)(y-y1)/h
where(x1,y1) is the point of intersection of the lines represented by the given equation.










