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Coordinate Geometry

Equation of bisectors

The equation of the bisectors of the angles between two intersecting lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ is derived using the locus property that any point on an angle bisector is equidistant from both lines. This topic covers acute versus obtuse bisector discrimination.

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Definition and statement

Locus Theorem for Angle Bisectors of Two Lines
The equations of the bisectors of the angles between the lines $L_1: A_1x + B_1y + C_1 = 0$ and $L_2: A_2x + B_2y + C_2 = 0$ are given by $$\frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}}$$

📐 Geometric Interpretation

Geometrically, the bisector represents the locus of a point moving such that the perpendicular distances from it to both intersecting lines are equal. This generates two mutually perpendicular straight lines dividing the four quadrants formed by the intersecting lines into symmetric angular halves.

⚙️ Algebraic & Structural Meaning

Algebraically, it unifies the combination of two linear equations into a pair of homogeneous or non-homogeneous second-degree equations, reflecting rotational symmetry and orthogonality of the bisecting system.

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 N/A Strictly Mandatory / Necessary If $A_1B_2 - A_2B_1 = 0$, the lines are parallel or coincident. Parallel lines do not intersect to form standard angles; instead, they have a single equidistant parallel line as their bisector.
Positive Constant Term ($C_1, C_2 > 0$) Sign convention for $C_1, C_2$ Essential for acute/obtuse discrimination If constant terms are not made positive before testing $A_1A_2 + B_1B_2$, the tests for acute and obtuse bisectors will yield inverted or false geometric conclusions.
Real coefficients $A_i, B_i, C_i \in \mathbb{R}$ Domain of coefficients Strictly Mandatory Imaginary or complex coefficients invalidate the perpendicular distance formula $\frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}$, destroying the geometric locus definition.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Bisector Containing Origin

$$A_1x + B_1y + C_1 = \pm \frac{\sqrt{A_1^2 + B_1^2}}{\sqrt{A_2^2 + B_2^2}} (A_2x + B_2y + C_2) \quad \text{with sign chosen such that constant terms have matching signs}$$

Rule: Make constant terms $C_1$ and $C_2$ positive. The bisector containing the origin uses the same sign (+ or -) that makes the constant terms positive on both sides after clearing radicals.

Application: Directly tested in UP PGT 2021 and DSSSB PGT Mathematics to find the bisector of the angle containing a specific point.
02

Acute vs. Obtuse Bisector Identification Test

$$ \text{Compute } \mathcal{S} = A_1A_2 + B_1B_2 $$

Rule: Make $C_1, C_2 > 0$. If $A_1A_2 + B_1B_2 < 0$, then the positive sign formula ($+$) gives the acute angle bisector. If $A_1A_2 + B_1B_2 > 0$, the positive sign gives the obtuse angle bisector.

03

Angle Between Bisectors

$$ \theta = 90^\circ (\frac{\pi}{2}) $$

Rule: The angle between any two angle bisectors of a pair of lines is always a right angle ($90^\circ$), regardless of the angle between the original lines.

04

Bisectors of Lines Through Origin ($C_1 = 0, C_2 = 0$)

$$ \frac{a_1x^2 + 2hxy + b_2y^2}{a_1 - b_1} = \frac{xy}{h} \implies h(x^2 - y^2) = (a - b)xy $$

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

05

Condition for Bisectors to Coincide with Coordinate Axes

$$ h = 0 \implies ax^2 + by^2 = 0 $$

Rule: If the cross-product coefficient $h = 0$ in the homogeneous equation, the coordinate axes themselves are the angle bisectors.

Application: Used for rapid elimination in matching-type questions.
06

Sum of Coefficients for Perpendicularity

$$ a + b = 0 $$

Rule: For a homogeneous pair $ax^2 + 2hxy + by^2 = 0$ representing lines, the sum of coefficients of $x^2$ and $y^2$ equalling zero ($a+b=0$) implies the lines are perpendicular.

Application: Essential speed-check rule for second-degree coordinate geometry items.

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 for acute/obtuse bisectors without first ensuring that $C_1 > 0$ and $C_2 > 0$.

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

Why it fails: If $C$ has a negative sign, the inequality test flips, yielding the obtuse bisector when acute was requested.

🛡️ Exam Defense: Always multiply the linear equation by $-1$ if its constant term $C$ is negative before running any angle discrimination test.
⚠️ COMMON ERROR #02

Confusing Acute and Obtuse Bisector Rules

Common Pitfall: Swapping the conditions when $A_1A_2 + B_1B_2$ is positive versus negative.

Counterexample: Assuming $A_1A_2 + B_1B_2 > 0$ gives acute.

Why it fails: Leads directly to selecting the wrong option among the two perpendicular bisector choices.

🛡️ Exam Defense: Memorize: If $A_1A_2 + B_1B_2 > 0$, the '+$' sign gives the *obtuse* bisector. Remember 'Plus is Obtuse when Product is Positive' (PO-PO).

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
General Bisector PairTwo intersecting lines $L_1, L_2$$\frac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}} = \pm \frac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}$Locus of points equidistant from both linesFind the equation of bisectors of given lines
Homogeneous Bisector PairLines passing through origin $ax^2+2hxy+by^2=0$$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$Bisectors of lines through originGiven second-degree homogeneous equation
Acute Angle Bisector$C_1 > 0, C_2 > 0$ and $A_1A_2 + B_1B_2 < 0$Use $+$ sign if $A_1A_2+B_1B_2 < 0$ (with $C_i>0$)Bisects the angle measuring $< 90^\circ$Keyword: 'acute angle bisector'
Obtuse Angle Bisector$C_1 > 0, C_2 > 0$ and $A_1A_2 + B_1B_2 > 0$Use $+$ sign if $A_1A_2+B_1B_2 > 0$ (with $C_i>0$)Bisects the angle measuring $> 90^\circ$Keyword: 'obtuse angle bisector'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

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

(A) $21x + 77y - 101 = 0$(B) $99x - 27y + 81 = 0$(C) $21x - 77y + 101 = 0$(D) $77x + 21y - 101 = 0$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Make constant terms positive. Rewrite second line as $-12x + 5y + 2 = 0$ or adjust. Here $C_1 = 7 > 0$, but $C_2 = -2 < 0$, so multiply second equation by $-1$: $-12x + 5y + 2 = 0$. Step 2: Compute $A_1A_2 + B_1B_2 = (3)(-12) + (-4)(5) = -36 - 20 = -56 < 0$. Step 3: Since product is negative, the acute bisector uses the $+$ sign: $\frac{3x - 4y + 7}{5} = +\frac{-12x + 5y + 2}{13}$. Step 4: Cross multiply and simplify to get $21x + 77y - 101 = 0$.
⚡ Method B: 25-Second Exam Shortcut
Check coefficient ratios and test $A_1A_2 + B_1B_2$ instantly to eliminate options with wrong signs.
💡 Tip: Always check $C_i > 0$ first. 80% of candidates miss this and mark the obtuse bisector.
CANONICAL PROBLEM 02 Dual-Method Breakdown

The angle bisectors of the pair of lines $x^2 - y^2 + 2pxy = 0$ are perpendicular to each other. What is the value of $p$?

(A) $0$(B) $1$(C) $-1$(D) Any real number
✓ Correct Answer: (D)
📚 Method A: Formal Textbook Derivation
Step 1: Identify coefficients from $ax^2 + 2hxy + by^2 = 0$ where $a = 1, h = p, b = -1$. Step 2: The bisectors are given by $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{p} \implies \frac{x^2 - y^2}{2} = \frac{xy}{p} \implies p(x^2 - y^2) - 2xy = 0$. Step 3: The angle bisectors are *always* perpendicular to each other regardless of the value of $p$. Thus $p$ can be any real number.
⚡ Method B: 25-Second Exam Shortcut
Recall the fundamental theorem that angle bisectors of any two lines are always mutually perpendicular ($90^\circ$), so no restriction on $p$ is required.
💡 Tip: A classic trap where students try to solve $a+b=0$ on the bisector equation instead of recalling that bisectors are always orthogonal.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: The equation of the bisectors of the angles between the coordinate axes is given by $...$
A $x^2 - y^2 = 0$
B $xy = 0$
C $x^2 + y^2 = 0$
D $x + y = 0$
Derivation & Analysis:

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

Q2
MCQ 2: If the lines $ax^2 + 2hxy + by^2 = 0$ are perpendicular, then their angle bisectors are given by $...$
A $h(x^2 - y^2) = (a - b)xy$
B $xy = 0$
C $ax^2 - by^2 = 0$
D $hx^2 - (a - b)xy - hy^2 = 0$
Derivation & Analysis:

When the original lines are perpendicular ($a + b = 0$), the angle bisectors coincide with the coordinate axes, represented by $xy = 0$.

Q3
MCQ 3: The angle between the two lines represented by $3x^2 + 10xy + 3y^2 = 0$ is bisected by the lines $...$
A $x^2 - y^2 = 0$
B $3x^2 - y^2 = 0$
C $xy = 0$
D $x^2 + y^2 = 1$
Derivation & Analysis:

Here $a = 3, h = 5, b = 3$. Using the bisector formula $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$, since $a - b = 3 - 3 = 0$, we get $x^2 - y^2 = 0$.

Q4
MCQ 4: For the lines $4x - 3y - 6 = 0$ and $12x + 5y - 2 = 0$, the value of $A_1A_2 + B_1B_2$ after making constant terms positive is $...$
A $48 - 15 = 33$
B $-48 - 15 = -63$
C $48 + 15 = 63$
D $-33$
Derivation & Analysis:

Make constants positive: $4x - 3y - 6 = 0 \implies -4x + 3y + 6 = 0$ ($A_1 = -4, B_1 = 3$), and $12x + 5y - 2 = 0 \implies -12x - 5y + 2 = 0$ ($A_2 = -12, B_2 = -5$). Then $A_1A_2 + B_1B_2 = (-4)(-12) + (3)(-5) = 48 - 15 = 33$. Wait, if both $C_1, C_2 > 0$ are achieved by multiplying by $-1$, $A_1A_2 + B_1B_2 = (-4)(-12) + (3)(-5) = 33 > 0$ (Obtuse).

Q5
MCQ 5: The bisector of the acute angle between the lines $x + y = 3$ and $x - y = 1$ is $...$
A $x = 2$
B $y = 1$
C $x - y = 0$
D $x + y = 2$
Derivation & Analysis:

Lines are $x + y - 3 = 0$ and $x - y - 1 = 0$. $A_1A_2 + B_1B_2 = (1)(1) + (1)(-1) = 0$. Since $A_1A_2 + B_1B_2 = 0$, the lines are perpendicular, and their bisectors form a 45-degree system. Testing point $(2,1)$ gives equal distance.

Q6
MCQ 6: Which of the following represents the product of perpendiculars from $(x_1, y_1)$ to the lines $ax^2 + 2hxy + by^2 = 0$?
A $\frac{ax_1^2 + 2hx_1y_1 + by_1^2}{(a-b)^2 + 4h^2}$
B $\frac{ax_1^2 + 2hx_1y_1 + by_1^2}{\sqrt{(a-b)^2 + 4h^2}}$
C $ax_1^2 + 2hx_1y_1 + by_1^2$
D 0
Derivation & Analysis:

Standard coordinate geometry result: the product of perpendiculars from a point $(x_1, y_1)$ to the pair of lines $ax^2 + 2hxy + by^2 = 0$ is $\frac{ax_1^2 + 2hx_1y_1 + by_1^2}{\sqrt{(a-b)^2 + 4h^2}}$.

Q7
MCQ 7: The combined equation of the bisectors of the angles between the lines given by $ax^2 + 2hxy + by^2 = 0$ is $...$
A $(a-b)(x^2 - y^2) + 4hxy = 0$
B $(a-b)(x^2 - y^2) - 4hxy = 0$
C $h(x^2 - y^2) = (a-b)xy$
D $hx^2 + (a-b)xy - hy^2 = 0$
Derivation & Analysis:

Cross-multiplying $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ yields $h(x^2 - y^2) = (a - b)xy$.

Q8
MCQ 8: Assertion (A): The angle bisectors of any two intersecting lines are perpendicular to each other. Reason (R): The product of the slopes of the two angle bisector lines is $-1$.
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 of (A)
C (A) is true but (R) is false
D (A) is false but (R) is true
Derivation & Analysis:

Angle bisectors always divide the angle into two equal parts summing to $90^\circ$, hence they are perpendicular, making their slope product $-1$.

Q9
MCQ 9: If the pair of lines $x^2 - 2pxy - y^2 = 0$ and $x^2 - 2qxy - y^2 = 0$ are such that each pair bisects the angle between the other pair, then $...$
A $pq = -1$
B $pq = 1$
C $p + q = 0$
D $p = q$
Derivation & Analysis:

For two homogeneous pairs to be mutual angle bisectors, the product of their $p$ and $q$ parameters must satisfy $pq = -1$.

Q10
MCQ 10: The distance between the origin and the point of intersection of the line bisectors of $2x^2 - 5xy + 2y^2 = 0$ is $...$
A $0$
B $1$
C Infinite
D Not defined
Derivation & Analysis:

All homogeneous equations of second degree $ax^2 + 2hxy + by^2 = 0$ and their bisectors pass through the origin $(0,0)$. Thus, the intersection point is the origin, and the distance is 0.

Common questions

Making $C_1$ and $C_2$ positive ensures that the origin lies in the acute or obtuse region defined by the standard inequality convention. Failing to do so inverts the sign of the test expression $A_1A_2 + B_1B_2$.

Rushing through the acute/obtuse bisector calculation without checking the sign of the constant terms or confusing the algebraic condition $A_1A_2 + B_1B_2 < 0$ for acute versus obtuse.

Memorize the direct homogeneous bisector formula $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ and use elimination based on the orthogonality ($90^\circ$) of bisectors to solve items.

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