We will learn how to find the equation of the bisector of the angle which contains the origin.
Algorithm to determine whether the origin lines in the obtuse angle or acute angle between the lines
Let the equation of the two lines be a\(_{1}\)x + b\(_{1}\)y + c\(_{1}\) = 0 and a\(_{2}\)x + b\(_{2}\)y + c\(_{2}\) = 0.
To determine whether the origin lines in the acute angles or obtuse angle between the lines we proceed as follows:
Step I: Obtain whether the constant terms c\(_{1}\) and c\(_{2}\) in the equations of the two lines are positive or not. Suppose not, make them positive by multiplying both sides of the equations by negative sign.
Step II: Determine the sign of a\(_{1}\)a\(_{2}\) + b\(_{1}\)b\(_{2}\).
Step III: If a\(_{1}\)a\(_{2}\) + b\(_{1}\)b\(_{2}\) > 0, then the origin lies in the obtuse angle and the “ + “ symbol gives the bisector of the obtuse angle. If a\(_{1}\)a\(_{2}\) + b\(_{1}\)b\(_{2}\) < 0, then the origin lies in the acute angle and the “ Positive (+) “ symbol gives the bisector of the acute angle i.e.,
\(\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}\) = + \(\frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}\)
Solved examples on the equation of the bisector of the angle which contains the origin:
1. Find the equations of the two bisectors of the angles between the straight lines 3x + 4y + 1 = 0 and 8x  6y  3 = 0. Which of the two bisectors bisects the angle containing the origin?
Solution:
3x + 4y + 1 = 0 ……….. (i)
8x  6y  3 = 0 ……….. (ii)
The equations of the two bisectors of the angles between the lines (i) and (ii)
\(\frac{3x + 4y + 1}{\sqrt{3^{2} + 4^{2}}}\) = + \(\frac{8x  6y  3}{\sqrt{8^{2} + (6)^{2}}}\)
⇒ 2 (3x + 4y + 1) = (8x  6y  3)
Therefore, the required two bisectors are given by,
6x + 8y + 2 = 8x+ 6y  3 (taking `+' sign)
⇒ 2x  14y = 5
And 6x+ 8y + 2 =  8x + 6y + 3 (taking `' sign)
⇒ 14x + 2y = 1
Since the constant terms in (i) and (ii) are of opposite signs, hence the bisector which bisects the angle containing the origin is
2 (3x + 4y + 1) =  (8x  6y  3)
⇒ 14x + 2y= 1.
2. For the straight lines 4x + 3y  6 = 0 and 5x + 12y + 9 = 0 find the equation of the bisector of the angle which contains the origin.
Solution:
To find the bisector of the angle between the lines which contains the origin, we first write down the equations of the given lines in such a form that the constant terms in the equations of the lines are positive. The equations of the given lines are
4x + 3y  6 = 0 ⇒ 4x  3y + 6 = 0 ……………………. (i)
5x + 12y + 9 = 0 ……………………. (ii)
Now the equation of the bisector of the angle between the lines which contains the origin is the bisector corresponding to the positive symbol i.e.,
\(\frac{4x  3y + 6}{\sqrt{(4)^{2} + (3)^{2}}}\) = + \(\frac{5x + 12y + 9}{\sqrt{5^{2} + 12^{2}}}\)
⇒ 52x – 39 y + 78 = 25x + 60y + 45
⇒ 7x + 9y – 3 = 0
Form (i) and (ii), we have a1a2 + b1b2 = 20 – 36 = 56 <0.
Therefore, the origin is situated in an acute angle region and the bisector of this angle is 7x + 9y – 3 = 0.
11 and 12 Grade Math
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