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Angle BisectorsTo find the equations of the bisectors of the angle between the lines a1x + b1y+ c1= 0 and a2x + b2y + c2= 0.A bisector is the locus of a point, which moves such that the perpendiculars drawn from it to the two given lines, are equal.The equations of the bisectors area1x+b1y+c1/va12+b12=+a2x+b2y+c2/va22+b22.AP is the bisector of an acute angle if,Tan (?PAN) = tan (?/2) is such that |tan ?/2| < 1.AP is an obtuse angle bisector if,Tan (?PAN) = tan (?/2) is such that |tan ?/2| > 1.Notes:•When both c1and c2are of the same sign, evaluate a1a2+ b1b2. If negative, then acute angle bisector is a1x+b1y+c1/va12+b12=+a2x+b2y+c2/va22+b22.•When both c1and c2are of the same sign, the equation of the bisector of the angle which contains the origin is a1x+b1y+c1/va12+b12=+a2x+b2y+c2/va22+b22.•Bisectors of the angle containing the point (a,ß) is a1x+b1y+c1/va12+b12=+a2x+b2y+c2/va22+b22if a1a+ b1ß+ c1and a2a+ b2ß+ c2have the same sign.•Bisectors of the angle containing the point (a, ß) is a1x+b1y+c1/va12+b12=+a2x+b2y+c2/va22+b22if a1a+ b1ß+ c1and a2a+ b2ß+ c2have the opposite sign.Illustration:For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0 , find the equation of the(i) bisector of the obtuse angle between them,(ii) bisector of the acute angle between them,(iii) bisector of the angle which contains (1, 2)Solution: Equations of bisectors of the angles between the given lines are 4x+3y–6/v42+32= + 5x+12y+9/v52+122? 9x – 7y – 41 = 0 and 7x + 9y – 3 = 0.If?is the angle between the line 4x + 3y – 6 = 0 and the bisector 9x – 7y – 41 = 0, then tan?= > 1.Hence(i) The bisector of the obtuse angle is 9x – 7y – 41 = 0.(ii) The bisector of the acute angle is 7x + 9y – 3 = 0.For the point (1, 2)4x + 3y – 6 = 4 × 1 + 3 × 2 – 6 > 0.5x + 12y + 9 = 12 × 12 + 9 > 0.Hence equation of the bisector of the angle containing the point (1, 2) is 4x+3y–6/5 = 5x+12y+9/13?9x – 7y – 41 = 0.Thanks & RegardsRinkoo GuptaAskIITians Faculty
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