How to write the equation of the bisector of the extreme angle
Writing the equation of the bisector of the student’s angle has been learned in the high school math program. In order to help them consolidate this extremely important Math knowledge, Le Hong Phong High School shared the following article. Here in addition to the theory to memorize, there are many more types of exercises, you can practice more.
I. HOW TO WRITE Equation of the bisector of the ANGLE
1. Solution method
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Given 2 intersecting lines: (d1) A1x + b1y + C1 = 0 and (d2): A2x + B2y + C2 = 0.
The equations of the bisectors of the angle formed by these two lines are:
Attention:
Given the line ∆: Ax + By + C = 0 and two points A(xA; yA); B(xB;yB ).
Let f(x;y) = Ax + By + C:
+ A and B are on the same side with respect to ∆ ⇔ f(xA; yA). f(xB; yB) > 0
+ A and B are on the opposite side of ∆ ⇔ f(xA; yA). f(xB; yB) < 0
2. Illustrated example
Example 1. Given a straight line d: 3x + 4y – 5 = 0 and 2 points A( 1; 3) ; B( 2; m) . Find m so that A and B are on the same side with respect to d?
A. m < 0 B. m > – 
C. m > 1 D. m = –
Solution guide
Two points A and B lie on either side of the line d if and only if:
( 3 + 12 – 5)( 6 + 4m – 5) < 0 or m > –
Choose B.
Example 2. In the plane with coordinate system Oxy, for triangle ABC there are A(0; 2) , B( 1; 2) and C(3; 6 ) . The equation of the internal bisector of angle A is:
A. 2x + y – 2 = 0 B. x – 2y + 4 = 0 C. 2x + y – 4 = 0 D. Other answers
The answer
+ We write the equation of the line AB and AC:
Line AB : 
( AB) : 0(x – 0) + 1(y – 2) = 0 or y – 2 = 0
AC line: 
( AC) : 4(x – 0) – 3(y – 2) = 0 or 4x – 3y + 6 = 0
Then the bisectors of angle A are: 
⇔ |4x – 3y + 6| = 5|y – 2|
Let f( x; y) = x – 2y + 4
⇒ f( B).f( C) =( 1 – 2.2 + 4) ( 3 – 2.6 + 4) = -5 < 0
⇒ B and C lie on the opposite side of the line: x – 2y + 4 = 0.
infer that the bisector in angle A is x – 2y + 4 = 0
Choose B.
Example 3. In the plane with coordinate system Oxy, for triangle ABC with A( 1; 5); B( -4; -5) and C( 4; -1) . The equation of the exterior bisector of angle A is:
A. y + 5 = 0 B. y – 5 = 0 C. x + 1 = 0 D. x – 1 = 0
The answer
+ We write the equation of the line AB and AC:
Line AB : 
⇒ ( AB) : 2( x – 1) – 1.( y – 5) = 0 or 2x – y + 3 = 0
AC line: 
( AC) : 2( x – 1) + 1( y – 5) = 0 or 2x + y – 7 = 0
Then the bisectors of angle A are: 
⇔ |2x – y + 3| = |2x + y – 7|
⇔ 
Let f( x; y) = y – 5
⇒ f(B).f(C) = ( -5 – 5).( -1 – 5) = 60 > 0
⇒ B and C are on the same side of the line: y – 5 = 0.
infer that the bisector outside angle A is y – 5 = 0
Choose B.
Example 4. In the plane with coordinate system Oxy, for two lines d1: 3x – 4y – 3 = 0 and d2 : 12x + 5y – 12 = 0. Equation of bisector of acute angle formed by two lines d1 and d2 is:
A. 3x + 11y – 3 = 0 B. 11x – 3y – 11 = 0 C. 3x – 11y – 3 =0 D. 11x + 3y – 11 = 0
The answer
The bisectors of the angles formed by two lines d1 and d2 are:
⇔ 13|3x – 4y – 3| = 5|12x + 5y – 12| .
Let I be the intersection of d1 and d2; coordinate I is the solution system
⇒ I( 1,0)
+ Call the line d: 3x + 11y – 3 = 0. Take the point M(-10; 3) on the line d.
Let H be the projection of M onto d1.
We have: IM = 
and
MH = d(M;d1 ) = 
= 9
I guess 
It follows that d: 3x + 11y – 3 = 0 is the obtuse angle bisector, so the acute angle bisector is 11x – 3y – 11 = 0.
Choose B.
II. EXERCISES ON DIGITAL LINE Equations
Exercise 1. Given two lines d: x + 2y + 3 = 0 and d’: 2x + y + 3 = 0. The equations of the bisectors of the angles formed by d and d’ are:
A. x + y = 0 and x – y + 4 = 0 . B. x – y + 4 = 0 and x + y – 2 = 0 .
C. x + y + 2 = 0 and x – y = 0 D. x + y + 1 = 0 and x – y – 3 = 0 .
Solution guide
The equations of the bisectors of the angles formed by d and d’ are:
So the equations of the bisectors formed by d and d’ are :
x – y = 0 and x + y + 2 = 0
Choose C.
Exercise 2. Which of the following pairs of lines is the bisector of the angles formed by two lines ∆1: x + 2y – 3 = 0 and ∆2: 2x – y + 3 = 0.
A. x + 3y – 2 = 0 and x = 3y. B. 3x = – y and x – 3y – 6 = 0.
C. 3x + y = 0 and –x + 3y – 6 = 0. D. Other answer
Solution guide
Let M(x; y) be the point on the bisector of two given lines.
⇒ d(M,Δ1) = d(M,Δ2) 
⇒ x + 2y – 3 = ± (2x – y + 3) 
Choose C.
Exercise 3: Let ABC be a triangle with AB: 2x – y + 4 = 0; AC: x – 2y – 6 = 0 and two points B; C belongs to Ox. The external bisector of angle BAC is
A. 3x – 3y – 2 = 0 B. x + y + 10 = 0 C. 3x + 3y + 2 = 0 D. x + y – 2 = 0
Solution guide:
Since two points B and C belong to Ox, the coordinates of those two points are: B(-2; 0) and C(6; 0).
Let M( x; y) be on the bisector of angle BAC
We have: d(M, AB) = d(M, AC) 
⇔ |2x – y + 4| = |x – 2y – 6|
⇔ 
+ Consider the position of two points B and C with respect to the line x + y + 10 = 0.
We have : (-2 + 0 + 10).( 6 + 0 + 10) > 0 so two points B and C lie on the same side of the line x + y + 10 = 0.
The line x + y + 10 = 0 is the exterior bisector of angle BAC.
Choose B.
Exercise 5: Given two lines d : x + 2y + 3 = 0 and d’ : 2x + y + 3 = 0. The equations of the bisectors of the angles formed by d and d’ are
A. x + y = 0 ; x – y + 2 = 0 B. x – y = 0 ; x + y + 2 = 0
C. x + y + 2 = 0 ; x – y = 0 D. x + y – 2 = 0 ; x – y – 1 = 0
Solution guide
Answer:
Reply:
We have: M(x ; y) is on the bisector if and only if :
d(M; d) = d(M; d’) 
⇔ |x + 2y + 3| = |2x + y + 3| ⇔ 
Exercise 6: Which of the following pairs of lines is the bisector of the angles formed by the line ∆: x + y = 0 and the horizontal axis Ox.
A. (1 + √2)x + y = 0 ; x – (1 – √2)y = 0 . B. (1 + √2)x + y = 0; x + (1 – √2)y = 0 .
C. (1 + √2)x – y = 0; x + (1 – √2)y = 0 . D. All wrong
Solution guide
Answer: EASY
Reply:
Let M(x; y) be the point on the bisector
According to the beginning of the lesson, we have: d( M; ∆) = d( M; Ox)
⇒ 
⇔ |x + y| = 2|y|
⇔
Lesson 7: Let ABC be a triangle with A(-2; -1); B(-1; 3) and C( 6; 1). Write the equation of the bisector of the exterior angle A of triangle ABC.
A. x – y + 1 = 0 B. 5x + 3y + 9 = 0 C. 3x + 3y – 5 = 0 D. x + y + 3 = 0
Solution guide
Answer: EASY
Reply:
+ Equation of line AB: 
⇒ Equation AB: 4( x + 2) – 1( y + 1) = 0 or 4x – y + 7 = 0
+ Equation of line AC: 
⇒ AC equation: 1( x + 2) – 4( y + 1) = 0 or x – 4y – 2 = 0
The equations of the bisectors of angle A are:
Set f1( x; y) = x + y + 3 we have: f1(B).f1(C) > 0
It follows that B and C are on the same side of d1 and different from d2.
So the equation of the bisector outside angle A is: x + y + 3 = 0.
Exercise 8: Let ABC be a triangle with the equations of sides AB: x + y – 1 = 0; AC: 7x – y + 2 = 0 and BC: 10x + y – 19 = 0. Write the equation of the bisector in the angle. A of triangle ABC.
A. 12x + 4y – 3 = 0 B. 2x + 6y + 7 = 0 C. 12x + 6y – 7 = 0 D. 2x – 6y +7 = 0
Solution guide
Answer: EASY
Reply:
+ AB and BC intersect at B, so the coordinates of B are solutions:
⇒ B( 2; -1)
+ AC and BC intersect at C, so the coordinate C is the solution:
⇒ C(1; 9)
The equations of the bisectors of angle A are:
Set f1( x; y) = 2x – 6y + 7 we get f1(B) . f1(C) < 0
It follows that B and C are on the opposite side of d1 and on the same side of d2.
So the equation of the bisector in angle A is: 2x – 6y + 7 = 0
Lesson 9: In the plane with coordinate system Oxy, let triangle ABC have A(2; -1) ; B( -1;3) and C( 4; -1) . The equation of the exterior bisector of angle A is:
A. y + 5 = 0 B. x + 2y = 0 C. x + 1 = 0 D. 2x – y – 5 = 0
Solution guide
Answer: REMOVE
Reply:
+ We write the equation of the line AB and AC:
Line AB : 
( AB) : 4(x – 2) + 3( y + 1) = 0 or 4x + 3y – 5 = 0
AC line: 
⇒ ( AC) : 0(x – 2) + 1( y + 1) = 0 or y + 1 = 0
Then the bisectors of angle A are: 
⇔ |4x + 3y – 5| = 5|y + 1|
⇔ 
Let f( x; y) = 2x – y – 5
⇒ f( B).f( C) = (- 2 – 3 – 5).( 2.4 + 1 – 5) = -40 < 0
⇒ B and C are on the opposite side of the line 2x – y – 5 = 0
infer that the bisector outside angle A is x + 2y = 0
Lesson 10: Let ABC be a triangle with A(1; -2); B( 2; 2) and C(5; -3). Write the equation of the internal bisector from the vertex A of triangle ABC?
A. 5x + 3y + 1 = 0 B. 3x + 2y + 1 = 0 C. x + y + 1 = 0 D. 5x + 2y – 1 = 0
Solution guide
Answer: A
Reply:
+ We have: AB = 
AC = 
⇒ AB = AC so triangle ABC is isosceles at A.
⇒ The internal bisector from A is also the median.
+ Let M be the midpoint of BC; M coordinates:
+ Line AM: 
⇒ AM equation: 5( x – 1) + 3( y + 2) = 0 or 5x + 3y + 1 = 0
Above, Le Hong Phong High School introduced to you how to write the equation of the bisector of an angle and many common types of exercises. Hopefully, this will be an essential resource to help you teach and learn better. See more how to find the direction vector of a line at this link!
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