Rinkoo Gupta
Last Activity: 10 Years ago
Angle Bisectors
To 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 are
a1x+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 & Regards
Rinkoo Gupta
AskIITians Faculty