Bisectors of the Angles between two given lines
Angular bisector is the locus of a point which moves in
such a way so that its distance from two intersecting lines remains same. .
The equations of the two bisectors of
the angles between the lines a1x + b1y + c1 =
0 and a2x + b2y + c2 = 0 are
If the two given lines are not
perpendicular i.e. a1 a2 + b1 b2 >
0, then one of these equations is the equation of the bisector of the acute
angle and the other that of the obtuse angle.
Note:
Whether both
lines are perpendicular or not but the angular bisectors of these lines will
always be mutually perpendicular.
The bisectors of the acute
and the obtuse angles:
Take one of the lines
and let its slope be m1 and take one of the bisectors and let its
slope be m2. If q be the acute angle between them,
then find
If
tanq > 1 then the bisector taken is the
bisector of the obtuse angle and the other one will be the bisector of the
acute angle.
If
0 < tanq < 1 then the bisector taken is
the bisector of the acute angle and the other one will be the bisector of the
obtuse angle.
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If
two lines are a1 x + b1y + c1 = 0 and a2x+
b2y + c2 = 0, then
=
will
represent the equation of the bisector of the acute or obtuse angle between the
lines according as c1c2(a1a2 + b1b2)
is negative or positive. .
The equation of the
bisector of the angle containing the origin:
Write the equations of
the two lines so that the constants c1 and c2 become
positive. Then the equation is
the equation of the bisector containing the origin.
Remark:
(i)
If a1a2
+ b1b2 < 0, then the origin will lie in the acute
angle and if
a1a2 + b1b2 > 0, then origin
will lie in the obtuse angle.
(ii)
The remark (i) is
helpful in finding the equation of bisector of the obtuse angle or acute angle
directly.
The equation of the
bisector of the angle which contains a given point:
The equation of the bisector of the
angle between the two lines containing the point (a,b)
is or
according
as
a1a + b1b
+ c1 and a2 a + b2b
+ c2 are of the same signs or of opposite signs.
Example 1
For
the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0, find the equation of
the
- bisector
of the obtuse angle between them.
- bisector
of the acute angle between them.
- bisector
of the angle which contains (1, 2).
Solution:
Equations of bisectors
of the angles between the given lines are
&⇒
9x – 7y – 41 = 0 and 7x + 9y – 3 = 0
If q
is the acute angle between the line 4x + 3y – 6 = 0 and the bisector 9x – 7y –
41 = 0, then tan q =
Hence
- The bisector of the obtuse angle is
9x – 7y – 41 = 0.
- The bisector of the acute angle is 7x
+ 9y – 3 = 0 The bisector of the angle containing the origin
&⇒
7x + 9y – 3 = 0. - For the point (1, 2), 4x + 3y – 6 =
4 ´ 1 + 3´2 – 6 > 0
5x + 12y + 9 = 12´2
+ 9 > 0
Hence equation of the bisector of the
angle containing the point
(1, 2) is
&⇒
9x – 7y – 41 = 0
Alternative:
Making C1 and C2
positive in the given equations, we get
–4x – 3y + 6 = 0 and 5x + 12y + 9 =
0
Since a1a2 + b1b2
= –20 –36 = –56 < 0, so the origin will lie in the acute angle. Hence
bisector of the acute angle is given by.
i.e. 9x – 7y – 41 = 0
Similarly bisector of obtuse angle is
7x + 9y – 3 = 0.
The equation of reflected
ray :
Let
L1 º a1x + b1y +
c1 = 0 be the incident ray in the line mirror L2 º
a2x + b2y + c2 = 0.
Let
L3 be the reflected ray from the line L2. Clearly L2
will be one of the bisectors of the angles between L1 and L3.
Since L3 passes through A, so .
L3
º L1 + lL2 = 0.
Let
(h, k) be a point on L2. Then, .
= .
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Since (h, k) lies on L2, a2h + b2k
+ c2 = 0
&⇒ a12
+ a22 l2 + 2a1a2l
+ b12 + b22l2 + 2b1b2l
= a12 + b12
&⇒ l
= 0 or
But l = 0 gives L3
= L1 . Hence L3 º .
Note:
Some
times the reflected ray L3 is also called the mirror image of L1
in L2 .