ANGLE BISECTOR :
EQUATION OF BISECTORS OF THE ANGLES BETWEEN TWO LINES :
`(i)` Equations of the bisectors of angles between the lines `ax + by+ c = 0 & a'x + b'y + c' = 0`
`(ab' ne a'b)` are ; `(ax+by+c)/(sqrt(a^2+b^2)) = - (a'x+b'y+c)/(sqrt(a'^2+b'^2))`
Explanation : As we know that angle bisector between two lines is the locus of the point which moves in
a plane such that its perpendicular distance from both the lines are equal.
Let `P(h, k)` be a moving point
`PM = PN`
` | (ah+bk+c)/(sqrt(a^2+b^2)) | = | (a'h+b'k+c')/(sqrt(a'^2+b'^2)) |`
`:.` Locus of `P(h, k)` is
`(ax+by+c)/(sqrt(a^2+b^2)) = - (a'x+b'y+c)/(sqrt(a'^2+b'^2))`
`(ii)` To discriminate between the acute angle bisector & the obtuse angle bisector
If `theta` be the angle between one of the lines & one of the bisectors, find tan `theta.`
If `|tan theta| < 1`, then `2 theta < 90^o` so that this bisector is the acute angle bisector.
If `|tan theta| > 1`, then we get the bisector to be the obtuse angle bisector.
`(iii)` To discriminate between the bisector of the angle containing the origin & that of the angle not containing
the origin . Rewrite the equations , `ax+ by+ c = 0 & a'x + b'y + c' = 0` such that the constant terms
`c, c'` are positive. Then ;
`(ax+by+c)/(sqrt(a^2+b^2)) = + (a'x+b'y+c)/(sqrt(a'^2+b'^2))` gives the equation of the bisector of the angle containing the origin
`(ax+by+c)/(sqrt(a^2+b^2)) = - (a'x+b'y+c)/(sqrt(a'^2+b'^2))` gives the equation of the bisector of the angle not containing the
`(iv)` To discriminate between acute angle bisector & obtuse angle bisector proceed as follows
Write `ax+ by+ c = 0 & a'x + b'y + c' = 0` such that constant tenns are positive.
If `aa'+ bb' < 0,` then the angle between the lines that contains the origin is acute and the equation of the
bisector of this acute angIe is
`(ax+by+c)/(sqrt(a^2+b^2)) = + (a'x+b'y+c)/(sqrt(a'^2+b'^2))` therefore `(ax+by+c)/(sqrt(a^2+b^2)) = + (a'x+b'y+c)/(sqrt(a'^2+b'^2))`
is the equation of other bisector.
If, however , `aa' + bb' > 0,` then the angle between the lines that contains the origin is obtuse & the
equation of the bisector of this obtuse angle is:
`(ax+by+c)/(sqrt(a^2+b^2)) = + (a'x+b'y+c)/(sqrt(a'^2+b'^2))` ; therefore `(ax+by+c)/(sqrt(a^2+b^2)) = + (a'x+b'y+c)/(sqrt(a'^2+b'^2))`
is the equation of other bisector.
`(v)` Another way of identifying an acute and obtuse angle bisector is as follows :
Let `L_1 = 0 & L_2 = 0` are the given lines `& u_1 = 0` and `u_2 = 0` are the
bisectors between `L_1 = 0 & L_2 = 0.` Take a point `P` on any one of the lines
`L_1 = 0 or L_2 = 0` and drop perpendicular on
`u_1 = 0 & u_2 = 0` as shown. If,
`|p| < |q| => u_1 ` is the acute angle bisector.
`|p| > |q| => u_1` is the obtuse angle bisector.
`|p| > |q| =>` the lines `L_1 & L_2` are perpendicular.
Note:-
Equation of straight lines passing through `P(x,. y 1) &` equally inclined with the lines
`a_1x + b_1y + c_1 = 0 & a_2x + b_2y + c_2= 0` are those which are parallel to the bisectors between these
two lines and passing through the point `P. `
`(vi)` Bisector in case of triangle
`CASE-I :` When vertices oftriangle are known.
`Method-1 :`
`1.` Find the length of sides of triangle `AB( c), BC(a)` and `CA(b)` by distance formula.
`2.` Find incentre lof triangle where `l = ((ax_1+bx_2+cx_3)/(a+b+c),(ay_1+by_2+cy_3)/(a+b+c))`
`3.` With the help ofincentre we can find all angle bisectors of triangle `ABC.`
`Method-2 :`
`1.` Find the point `D` which divides the side `BC` in ratio `c : b.` and also find the points `E` and `F`
similarly.
`2.` Find the equations of `AD, BE` and `CF.`
`CASE-II :` When the equation of sides are given.
`Method-I :`
`1.` Compute `tanA , tanB , tanC` and arrange the lines in descending order of their slopes.
`2.` With the help of angles we can find all acute//obtuse angle bisectors.
`Method-2 :`
`1.` Plot the lines approximately and compute bisectors containing or not containing the origin.
`(i)` Equations of the bisectors of angles between the lines `ax + by+ c = 0 & a'x + b'y + c' = 0`
`(ab' ne a'b)` are ; `(ax+by+c)/(sqrt(a^2+b^2)) = - (a'x+b'y+c)/(sqrt(a'^2+b'^2))`
Explanation : As we know that angle bisector between two lines is the locus of the point which moves in
a plane such that its perpendicular distance from both the lines are equal.
Let `P(h, k)` be a moving point
`PM = PN`
` | (ah+bk+c)/(sqrt(a^2+b^2)) | = | (a'h+b'k+c')/(sqrt(a'^2+b'^2)) |`
`:.` Locus of `P(h, k)` is
`(ax+by+c)/(sqrt(a^2+b^2)) = - (a'x+b'y+c)/(sqrt(a'^2+b'^2))`
`(ii)` To discriminate between the acute angle bisector & the obtuse angle bisector
If `theta` be the angle between one of the lines & one of the bisectors, find tan `theta.`
If `|tan theta| < 1`, then `2 theta < 90^o` so that this bisector is the acute angle bisector.
If `|tan theta| > 1`, then we get the bisector to be the obtuse angle bisector.
`(iii)` To discriminate between the bisector of the angle containing the origin & that of the angle not containing
the origin . Rewrite the equations , `ax+ by+ c = 0 & a'x + b'y + c' = 0` such that the constant terms
`c, c'` are positive. Then ;
`(ax+by+c)/(sqrt(a^2+b^2)) = + (a'x+b'y+c)/(sqrt(a'^2+b'^2))` gives the equation of the bisector of the angle containing the origin
`(ax+by+c)/(sqrt(a^2+b^2)) = - (a'x+b'y+c)/(sqrt(a'^2+b'^2))` gives the equation of the bisector of the angle not containing the
`(iv)` To discriminate between acute angle bisector & obtuse angle bisector proceed as follows
Write `ax+ by+ c = 0 & a'x + b'y + c' = 0` such that constant tenns are positive.
If `aa'+ bb' < 0,` then the angle between the lines that contains the origin is acute and the equation of the
bisector of this acute angIe is
`(ax+by+c)/(sqrt(a^2+b^2)) = + (a'x+b'y+c)/(sqrt(a'^2+b'^2))` therefore `(ax+by+c)/(sqrt(a^2+b^2)) = + (a'x+b'y+c)/(sqrt(a'^2+b'^2))`
is the equation of other bisector.
If, however , `aa' + bb' > 0,` then the angle between the lines that contains the origin is obtuse & the
equation of the bisector of this obtuse angle is:
`(ax+by+c)/(sqrt(a^2+b^2)) = + (a'x+b'y+c)/(sqrt(a'^2+b'^2))` ; therefore `(ax+by+c)/(sqrt(a^2+b^2)) = + (a'x+b'y+c)/(sqrt(a'^2+b'^2))`
is the equation of other bisector.
`(v)` Another way of identifying an acute and obtuse angle bisector is as follows :
Let `L_1 = 0 & L_2 = 0` are the given lines `& u_1 = 0` and `u_2 = 0` are the
bisectors between `L_1 = 0 & L_2 = 0.` Take a point `P` on any one of the lines
`L_1 = 0 or L_2 = 0` and drop perpendicular on
`u_1 = 0 & u_2 = 0` as shown. If,
`|p| < |q| => u_1 ` is the acute angle bisector.
`|p| > |q| => u_1` is the obtuse angle bisector.
`|p| > |q| =>` the lines `L_1 & L_2` are perpendicular.
Note:-
Equation of straight lines passing through `P(x,. y 1) &` equally inclined with the lines
`a_1x + b_1y + c_1 = 0 & a_2x + b_2y + c_2= 0` are those which are parallel to the bisectors between these
two lines and passing through the point `P. `
`(vi)` Bisector in case of triangle
`CASE-I :` When vertices oftriangle are known.
`Method-1 :`
`1.` Find the length of sides of triangle `AB( c), BC(a)` and `CA(b)` by distance formula.
`2.` Find incentre lof triangle where `l = ((ax_1+bx_2+cx_3)/(a+b+c),(ay_1+by_2+cy_3)/(a+b+c))`
`3.` With the help ofincentre we can find all angle bisectors of triangle `ABC.`
`Method-2 :`
`1.` Find the point `D` which divides the side `BC` in ratio `c : b.` and also find the points `E` and `F`
similarly.
`2.` Find the equations of `AD, BE` and `CF.`
`CASE-II :` When the equation of sides are given.
`Method-I :`
`1.` Compute `tanA , tanB , tanC` and arrange the lines in descending order of their slopes.
`2.` With the help of angles we can find all acute//obtuse angle bisectors.
`Method-2 :`
`1.` Plot the lines approximately and compute bisectors containing or not containing the origin.
