Coordinate Geometry

Angle Between Angle Bisectors

The angle between the angle bisectors of two intersecting lines in a coordinate plane is always a right angle ($90^\circ$), reflecting the orthogonal property of internal and external bisectors.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Orthogonality of Internal and External Angle Bisectors

Let two intersecting lines be given by $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$. The equations of their angle bisectors are given by \frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}. The angle between these two bisector lines is always $\frac{\pi}{2}$ radians ($90^\circ$).

📐 Geometric Interpretation

Geometrically, any point on an angle bisector is equidistant from the two original lines. The internal and external bisectors divide the supplementary angles formed at the intersection of two lines into equal halves, creating a net perpendicular framework (orthogonal grid) at the vertex.

⚙️ Algebraic & Structural Meaning

Algebraically, the product of the slopes of the two angle bisector lines is $-1$ (i.e., $m_1 m_2 = -1$), which confirms their perpendicularity regardless of the angle between the original lines.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Non-parallel lines condition	$a_1b_2 - a_2b_1 \neq 0$	Strictly Mandatory / Necessary	If lines are parallel ($a_1b_2 - a_2b_1 = 0$), the intersection point lies at infinity, and standard angle bisectors degenerate into parallel equidistant lines.
Non-zero normal vectors	`$a_1^2 + b_1^2 \neq 0$ and $a_2^2 + b_2^2 \neq 0$`	Strictly Mandatory	If either equation reduces to a trivial identity ($0 = 0$), the line equation is undefined, rendering distance formulas invalid.
Positive constant term standardization	`$c_1, c_2$ adjustment such that $c_1, c_2 > 0$ or matching signs of $a_1a_2 + b_1b_2$`	Essential for acute/obtuse bisector identification	Failing to make constant terms positive before applying the positive/negative sign test leads to incorrectly identifying the obtuse bisector as the acute one.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Direct Orthogonality Test

$$m_{b_1} \cdot m_{b_2} = -1$$

Rule: Never calculate slopes of bisectors individually to check perpendicularity; the angle between any pair of angle bisectors of two intersecting lines is universally $90^\circ$.

Application: undefined

Acute vs. Obtuse Bisector Test

$$a_1a_2 + b_1b_2$$

Rule: Make $c_1, c_2 > 0$. If $a_1a_2 + b_1b_2 > 0$, then the '$-$' sign gives the acute angle bisector, and the '$+$' sign gives the obtuse angle bisector.

Application: High frequency in DSSSB PGT and KVS PGT for identifying acute/obtuse region bisectors.

Bisector Containing Origin

$$c_1, c_2 \text{ signs determine origin region}$$

Rule: The bisector that contains the origin $(0,0)$ uses the same sign as the constant terms when $c_1, c_2$ have matching signs after making constant terms positive.

Combined Equation of Bisectors

$$\frac{(a_1x + b_1y + c_1)^2}{a_1^2 + b_1^2} = \frac{(a_2x + b_2y + c_2)^2}{a_2^2 + b_2^2}$$

Rule: Use the squared difference formulation to directly generate the homogeneous/non-homogeneous combined pair of bisectors without splitting into separate equations.

Bisector of Lines through Origin

$$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$$

Rule: For homogeneous lines $ax^2 + 2hxy + by^2 = 0$, the angle bisectors equation is given instantly by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.

Application: Direct formula recall in UP PGT.

Rapid Elimination via Origin Test

$$L_1(0,0) \cdot L_2(0,0) vs \text{ Bisector Plug-in}$$

Rule: Substitute $(0,0)$ into the given options to instantly eliminate options that do not satisfy the required region property.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Forgetting to Make Constant Terms Positive

Common Pitfall: Applying the $a_1a_2 + b_1b_2$ test without first ensuring that $c_1$ and $c_2$ are positive.

Counterexample: $3x - 4y - 5 = 0$ and $5x + 12y - 10 = 0$

Why it fails: If constant terms have differing or unstandardized signs, the test for acute and obtuse bisectors gets inverted, leading to a guaranteed wrong option.

🛡️ Exam Defense: Always multiply the equation by $-1$ if the constant term $c$ is negative before testing acute/obtuse bisectors.

⚠️ COMMON ERROR #02

Confusing Acute Bisector with Angle Between Lines

Common Pitfall: Confusing the angle between the two original lines with the angle between their bisectors ($90^\circ$).

Counterexample: \text{Angle between bisectors is always } 90^\circ \text{ regardless of lines}

Why it fails: Candidates calculate $\tan \theta = \pm \frac{m_1 - m_2}{1 + m_1m_2}$ unnecessarily when asked for the angle between the bisectors themselves.

🛡️ Exam Defense: Recognize that the angle between the internal and external bisectors is universally a right angle ($90^\circ$).

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Angle Between Two Lines	$1 + m_1m_2 \neq 0$	$\tan \theta = \left\| \frac{m_1 - m_2}{1 + m_1m_2} \right\|$	Acute/obtuse angle formed by intersecting lines	Asked for angle between $L_1$ and $L_2$
Angle Bisectors Equation	$a_1b_2 - a_2b_1 \neq 0$	$\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$	Locus of points equidistant from both lines	Asked for equation of bisectors
Angle Between Bisectors	Lines intersect non-perpendicularly	$\theta = 90^\circ$ ($m_{b_1}m_{b_2} = -1$)	Bisectors are always orthogonal	Asked for angle between bisector pair
Homogeneous Bisectors	$h^2 - ab > 0$	$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$	Bisectors of lines passing through origin	Joint equation of lines given

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the angle between the angle bisectors of the lines $3x - 4y + 7 = 0$ and $12x + 5y - 2 = 0$.

(A) $0^\circ$(B) $45^\circ$(C) $90^\circ$(D) $180^\circ$

📚 Method A: Formal Textbook Derivation

Step 1: Identify slopes of the two lines. Step 2: Find bisector equations using standard formula. Step 3: Calculate slopes of bisectors $m_1'$ and $m_2'$. Step 4: Verify $m_1' m_2' = -1$, yielding $90^\circ$.

⚡ Method B: 25-Second Exam Shortcut

Recognize instantly that the angle between any pair of angle bisectors of two intersecting straight lines is universally $90^\circ$. Mark (C) in 5 seconds.

💡 Tip: Do not waste time calculating bisector equations when asked for the angle between them.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The acute angle bisector of the lines $x + y - 3 = 0$ and $7x - y + 5 = 0$ is:

(A) $3x + 5y - 7 = 0$(B) $3x - 5y + 4 = 0$(C) $2x + 3y - 1 = 0$(D) $x - 2y + 3 = 0$

📚 Method A: Formal Textbook Derivation

Step 1: Check $c_1 = -3$ (make positive: $-x - y + 3 = 0$) and $c_2 = 5$. Step 2: Compute $a_1a_2 + b_1b_2 = (1)(7) + (1)(-1) = 6 > 0$. Step 3: Since $a_1a_2 + b_1b_2 > 0$, the acute bisector is given by the minus sign. Step 4: Simplify to get $3x + 5y - 7 = 0$.

⚡ Method B: 25-Second Exam Shortcut

Test point substitution or verify product $a_1a_2 + b_1b_2$ sign to pick the correct sign instantly.

💡 Tip: Always standardize constant terms to be positive before applying the acute/obtuse test.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: What is the angle between the internal and external angle bisectors of two intersecting straight lines?

A $0^\circ$

B $30^\circ$

C $45^\circ$

D $90^\circ$

Derivation & Analysis:

The internal and external bisectors of any two intersecting lines are always perpendicular to each other, so the angle between them is $90^\circ$.

MCQ 2: The combined equation of the angle bisectors of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by:

A $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$

B $\frac{x^2 - y^2}{a + b} = \frac{xy}{h}$

C $\frac{x^2 + y^2}{a - b} = \frac{xy}{h}$

D $\frac{x^2 - y^2}{h} = \frac{xy}{a - b}$

Derivation & Analysis:

Standard textbook formula for the angle bisectors of homogeneous second-degree equations is $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.

MCQ 3: If the lines $ax + by + c = 0$ and $a'x + b'y + c' = 0$ are perpendicular, then the angle between their angle bisectors is:

A $0^\circ$

B $45^\circ$

C $90^\circ$

D $180^\circ$

Derivation & Analysis:

Regardless of whether the original lines are perpendicular, parallel, or acute-angled, the angle between their angle bisectors is always $90^\circ$.

MCQ 4: For the lines $3x - 4y + 1 = 0$ and $4x + 3y - 7 = 0$, the value of $a_1a_2 + b_1b_2$ is:

A $0$

B $12$

C $-12$

D $25$

Derivation & Analysis:

Here $a_1 = 3, b_1 = -4$ and $a_2 = 4, b_2 = 3$. Then $a_1a_2 + b_1b_2 = (3)(4) + (-4)(3) = 12 - 12 = 0$. This indicates the original lines are perpendicular.

MCQ 5: Which sign gives the acute angle bisector for $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ when $c_1, c_2 > 0$ and $a_1a_2 + b_1b_2 > 0$?

A Plus sign ($+$)

B Minus sign ($-$)

C Both signs give acute

D Neither sign

Derivation & Analysis:

When $c_1, c_2 > 0$ and $a_1a_2 + b_1b_2 > 0$, the minus sign yields the acute angle bisector.

MCQ 6: The equation of the bisectors of the angle between the coordinate axes ($x = 0$ and $y = 0$) is:

A $x^2 - y^2 = 0$

B $x^2 + y^2 = 0$

C $xy = 0$

D $x + y = 0$

Derivation & Analysis:

The coordinate axes are $x = 0$ and $y = 0$. Their bisectors are $y = \pm x$, which combine to give $x^2 - y^2 = 0$.

MCQ 7: If the lines $x \cos \alpha_1 + y \sin \alpha_1 = p_1$ and $x \cos \alpha_2 + y \sin \alpha_2 = p_2$ intersect, their angle bisectors are obtained by using:

A Normal form equality with $\pm$

B Slope-intercept form

C Parametric form

D Determinant expansion

Derivation & Analysis:

In normal form, since $\sqrt{a^2+b^2} = 1$, the bisectors are simply given by $x \cos \alpha_1 + y \sin \alpha_1 - p_1 = \pm (x \cos \alpha_2 + y \sin \alpha_2 - p_2)$.

MCQ 8: Assertion (A): The angle between the angle bisectors of two intersecting lines is $90^\circ$. Reason (R): The internal and external bisectors divide the angle between lines into parts summing to $180^\circ$, hence bisecting them orthogonally.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Both statements are mathematically rigorous and directly explain the orthogonal property of angle bisectors.

MCQ 9: Find the combined equation of the angle bisectors of the pair of lines $x^2 - y^2 = 0$.

A $xy = 0$

B $x^2 + y^2 = 0$

C $x^2 - y^2 = 0$

D $(x+y)^2 = 0$

Derivation & Analysis:

For $x^2 - y^2 = 0$, $a = 1, h = 0, b = -1$. Using $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ gives $h=0$ in denominator, indicating bisectors are $xy = 0$ (the coordinate axes themselves).

Q10

MCQ 10: If $ax^2 + 2hxy + by^2 = 0$ represents perpendicular lines, then what can be said about their angle bisectors?

A The bisectors are perpendicular to each other

B They coincide with the original lines

C They are always at $45^\circ$ to the coordinate axes

D They are parallel

Derivation & Analysis:

The two angle bisectors bisect a pair of perpendicular lines. They lie along the bisectors of the four right angles, so the angle between the two bisectors is also $90^\circ$.

Common questions

No. Most questions only test whether you know that the angle between the two bisectors is $90^\circ$ or require checking the sign of $a_1a_2 + b_1b_2$ for quick identification.

Forgetting to ensure that the constant terms $c_1$ and $c_2$ are positive before applying the acute/obtuse test, which inverts the sign rule.

Use the substitution shortcut: test options against given points or check the homogeneity and coefficient relations like $a+b=0$ to eliminate incorrect choices instantly.

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