Equations of the Bisectors of the Angles between Two Straight Lines

We will learn how to find the equations of the bisectors of the angles between two straight lines.

Prove that the equation of the bisectors of the angles between the lines a\(_{1}\)x + b\(_{1}\)y + c\(_{1}\) = 0 and a\(_{2}\)x + b\(_{2}\)y + c\(_{2}\) = 0 are given by \(\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}\) = ±\(\frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}\).

Let us assume the two given straight lines be PQ and RS whose equations are a\(_{1}\)x + b\(_{1}\)y + c\(_{1}\) = 0 and a\(_{2}\)x + b\(_{2}\)y + c\(_{2}\) = 0 respectively, where c\(_{1}\) and c\(_{2}\) are of the same symbols.

First we will find the equations of the bisectors of the angles between the lines a\(_{1}\)x + b\(_{1}\)y + c\(_{1}\) = 0 and a\(_{2}\)x + b\(_{2}\)y + c\(_{2}\) = 0.

Now, let us assume that the two straight lines PQ and RS intersect at T and ∠PTR contains origin O.

Again, let us assume that TU is the bisector of ∠PTR and Z(h, k) is any point on TU. Then the origin O and the point Z are on the same side of both the lines PQ and RS.

Therefore, c\(_{1}\), and (a\(_{1}\)h + b\(_{1}\)k + c\(_{1}\)) are of the same symbols and c\(_{2}\) and (a\(_{2}\)h + b\(_{2}\)k + c\(_{2}\)) are also of the same symbols.

Since, we already assumed that c\(_{1}\), and c\(_{2}\), are of the same symbols, thus, (a\(_{1}\)h + b\(_{1}\)k + c\(_{1}\)) and (a\(_{2}\)h + b\(_{2}\)k + c\(_{2}\)) shall be of the same symbols.

Therefore, the lengths of the perpendiculars from Z upon PQ and RS are of the same symbols. Now, if ZA ⊥ PQ and ZB ⊥ RS then it implies that ZA = ZB.

⇒ \(\frac{a_{1}h + b_{1}k + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}\) = \(\frac{a_{2}h + b_{2}k + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}\)

Therefore, the equation to the locus of Z (h, k) is,

\(\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}\) = \(\frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}\)………… (i), which is the equation of the bisector of the angle containing the origin.


Algorithm to find the bisector of the angle containing the origin:

Let the equations 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 find the bisector of the angle containing the origin, we proceed as follows:

Step I: First check whether the constant terms c\(_{1}\) and c\(_{2}\) in the given equations of two straight lines are positive or not. Suppose not, then multiply both the sides of the equations by -1 to make the constant term positive.

Step II: Now obtain the bisector corresponding to the positive symbol i.e. 

\(\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}\) = + \(\frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}\), which is the required bisector of the angle containing the origin.

Note:

The bisector of the angle containing the origin means the bisector of that angle between the two straight lines which contains the origin within it.

Again, ∠QTR does not contain the origin. Suppose, TV be the bisector of ∠QTR and Z'(α, β) be any point on TV then the origin O and Z' are on the same side of the straight line (PQ) but they are on opposite sides of the straight line RS.

Therefore, c\(_{1}\) and (a\(_{1}\)α + b\(_{1}\)β + c\(_{1}\)) are of the same symbols but c\(_{2}\) and (a\(_{2}\)α + b\(_{2}\)β + c\(_{2}\)), are of opposite symbols.

Since, we already assumed that, c\(_{1}\), and c\(_{2}\), are of the same symbols, thus, (a\(_{1}\)α + b\(_{1}\)β + c\(_{1}\)) and (a\(_{2}\)α + b\(_{2}\)β + c\(_{2}\)) shall be of opposite symbols.

Therefore, the lengths of the perpendiculars from Z' upon PQ and RS are of opposite symbols. Now, if Z'W ⊥ PQ and Z'C ⊥ RS then it readily follows that Z'W = -Z'C

⇒ \(\frac{a_{1}α + b_{1}β + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}\) = - \(\frac{a_{2}α + b_{2}β + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}\) 

Therefore, the equation to the locus of Z' (α, β) is

\(\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}\) = - \(\frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}\) ………… (ii), which is the equation of the bisector of the angle not containing the origin.

From (i) and (ii) it is seen that the equations of the bisectors of the angles between the lines a\(_{1}\)x + b\(_{1}\)y + c\(_{1}\) = 0 and a\(_{2}\)x + b\(_{2}\)y + c\(_{2}\) = 0 are \(\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}\) = ±\(\frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}\).

Note: The bisectors (i) and (ii) are perpendicular to each other.


Algorithm to find the bisectors of acute and obtuse angles between two lines:

Let the equations 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 separate the bisectors of the obtuse and acute angles between the lines we proceed as follows:

Step I: First check whether the constant terms c\(_{1}\) and c\(_{2}\) in the two equations are positive or not. Suppose not, then multiply both the sides of the given equations by -1 to make the constant terms positive.

Step II: Determine the symbols of the expression a\(_{1}\)a\(_{2}\) + b\(_{1}\)b\(_{2}\).

Step III: If a\(_{1}\)a\(_{2}\) + b\(_{1}\)b\(_{2}\) > 0, then the bisector corresponding to “ + “ symbol gives the obtuse angle bisector and the bisector corresponding to “ - “ is the bisector of the acute angle between the lines i.e.

\(\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}\) = + \(\frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}\) and \(\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}\) = - \(\frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}\)

are the bisectors of obtuse and acute angles respectively.

If a\(_{1}\)a\(_{2}\) + b\(_{1}\)b\(_{2}\) < 0, then the bisector corresponding to “ + “ and “ - “ symbol give the acute and obtuse angle bisectors respectively i.e.

\(\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}\) = + \(\frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}\) and \(\frac{a_{1}x + b_{1}y + c_{1}}{\sqrt{a_{1}^{2} + b_{1}^{2}}}\) = - \(\frac{a_{2}x + b_{2}y + c_{2}}{\sqrt{a_{2}^{2} + b_{2}^{2}}}\)

are the bisectors of acute and obtuse angles respectively.


Solved examples to find the equations of the bisectors of the angles between two given straight lines:

1. Find the equations of the bisectors of the angles between the straight lines 4x - 3y + 4 = 0 and 6x + 8y - 9 = 0.

Solution:

The equations of the bisectors of the angles between 4x - 3y + 4 = 0 and 6x + 8y - 9 = 0 are

\(\frac{4x - 3y + 4}{\sqrt{4^2} + (-3)^{2}}\) = ± \(\frac{6x + 8y - 9}{\sqrt{6^2} + 8^{2}}\)

⇒ \(\frac{4x - 3y + 4}{5}\) = ±\(\frac{6x + 8y - 9}{10}\)

⇒ 40x - 30y + 40 = ±(30x + 40y - 45)

Taking positive sign, we get,

⇒ 40x - 30y + 40 = +(30x + 40y - 45)

⇒ 2x - 14y + 17 = 0

Taking negative sign, we get,

⇒ 40x - 30y + 40 = -(30x + 40y - 45)

⇒ 40x - 30y + 40 = -30x - 40y + 45

⇒ 70x + 10y - 5 = 0

Therefore the equations of the bisectors of the angles between the straight lines 4x - 3y + 4 = 0 and 6x + 8y - 9 = 0 are 2x - 14y + 17 = 0 and 70x + 10y - 5 = 0.


2. Find the equation of the obtuse angle bisector of lines 4x - 3y + 10 = 0 and 8y - 6x - 5 = 0.

Solution:

First we make the constant terms positive in the given two equations.

Making positive terms positive, the two equations becomes

4x - 3y + 10 = 0 and 6x - 8y + 5 = 0

Now, a\(_{1}\)a\(_{2}\) + b\(_{1}\)b\(_{2}\) = 4 × 6 + (-3) × (-8) = 24 + 24 = 48, which is positive. Hence, “+” symbol gives the obtuse angle bisector. The obtuse angle bisector is

⇒ \(\frac{4x - 3y + 10}{\sqrt{4^2} + (-3)^{2}}\) = + \(\frac{6x - 8y + 5}{\sqrt{6^2} + (-8)^{2}}\)

⇒ \(\frac{4x - 3y + 10}{5}\) = +\(\frac{6x - 8y + 5}{10}\)

⇒ 40x - 30y + 100 = 30x - 40y - 50

⇒ 10x + 10y + 150 = 0

x + y + 15 = 0, which is the required obtuse angle bisector.

 The Straight Line




11 and 12 Grade Math

From Equations of the Bisectors of the Angles between Two Straight Lines to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Roman Numerals | System of Numbers | Symbol of Roman Numerals |Numbers

    Feb 22, 24 04:21 PM

    List of Roman Numerals Chart
    How to read and write roman numerals? Hundreds of year ago, the Romans had a system of numbers which had only seven symbols. Each symbol had a different value and there was no symbol for 0. The symbol…

    Read More

  2. Worksheet on Roman Numerals |Roman Numerals|Symbols for Roman Numerals

    Feb 22, 24 04:15 PM

    Roman Numbers Table
    Practice the worksheet on roman numerals or numbers. This sheet will encourage the students to practice about the symbols for roman numerals and their values. Write the number for the following: (a) V…

    Read More

  3. Roman Symbols | What are Roman Numbers? | Roman Numeration System

    Feb 22, 24 02:30 PM

    Roman Numbers
    Do we know from where Roman symbols came? In Rome, people wanted to use their own symbols to express various numbers. These symbols, used by Romans, are known as Roman symbols, Romans used only seven…

    Read More

  4. Place Value | Place, Place Value and Face Value | Grouping the Digits

    Feb 19, 24 11:57 PM

    Place-value of a Digit
    The place value of a digit in a number is the value it holds to be at the place in the number. We know about the place value and face value of a digit and we will learn about it in details. We know th…

    Read More

  5. Math Questions Answers | Solved Math Questions and Answers | Free Math

    Feb 19, 24 11:14 PM

    Math Questions Answers
    In math questions answers each questions are solved with explanation. The questions are based from different topics. Care has been taken to solve the questions in such a way that students

    Read More