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Coordinate Geometry

Homogeneous equation of second degree and pair of straight lines

The homogeneous equation of second degree in two variables, $ax^2 + 2hxy + by^2 = 0$, universally represents a pair of straight lines passing through the origin.

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

Homogeneity and Joint Equation Principle for Pair of Lines Through Origin
A homogeneous equation of second degree in $x$ and $y$, given by $$ax^2 + 2hxy + by^2 = 0,$$ always represents two real or imaginary straight lines passing through the origin $(0,0)$. The separate equations of these lines are $y = m_1x$ and $y = m_2x$, where $m_1$ and $m_2$ are the roots of the auxiliary quadratic equation $bm^2 + 2hm + a = 0$.

📐 Geometric Interpretation

Geometrically, every term in the equation is of degree two, ensuring that if $(x,y)$ lies on the locus, then $(\lambda x, \lambda y)$ also lies on the locus for any scalar $\lambda$. This scaling invariance forces the locus to be a degenerate conic consisting of two intersecting straight lines anchored at the origin.

⚙️ Algebraic & Structural Meaning

Algebraically, it establishes a bijective correspondence between the factorization of quadratic forms into linear factors over $\mathbb{R}$ (or $\mathbb{C}$) and the geometric configuration of concurrent lines through the origin.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Degree Homogeneity Polynomial equation where every non-zero term has total degree 2 Strictly Mandatory / Necessary If linear or constant terms are present ($gx + fy + c \neq 0$), the lines are shifted away from the origin, requiring the general second-degree equation.
Discriminant for Line Reality ($h^2 \ge ab$) Real domain $a, h, b \in \mathbb{R}$ Strictly Mandatory for Real Lines If $h^2 < ab$, the auxiliary roots are complex conjugates, yielding a pair of imaginary lines intersecting only at the real point $(0,0)$.
Non-degeneracy / Non-triviality ($a^2 + h^2 + b^2 > 0$) Coefficient space Essential for non-zero locus If $a = h = b = 0$, the equation reduces to $0 = 0$, representing the entire Cartesian plane $\mathbb{R}^2$ rather than a pair of lines.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Angle Between Pair of Lines Through Origin

$$ \tan\theta = \pm \frac{2\sqrt{h^2 - ab}}{a + b} $$

Rule: Direct formula to find the acute angle $\theta$ between the lines $ax^2 + 2hxy + by^2 = 0$.

Application: Directly tested in UP PGT/TGT and KVS PGT to calculate angle or find unknown parameters when lines are perpendicular or coincident.
02

Condition for Perpendicular Lines

$$ a + b = 0 $$

Rule: The lines are mutually perpendicular if and only if the sum of the coefficients of $x^2$ and $y^2$ equals zero.

03

Condition for Coincident (Identical) Lines

$$ h^2 = ab $$

Rule: The lines coincide if the discriminant of the auxiliary quadratic equation vanishes.

Application: Standard question pattern in DSSSB and LT Grade assessing degenerate conics.
04

Condition for Real and Distinct Lines

$$ h^2 > ab $$

Rule: The lines are distinct and real if the discriminant $h^2 - ab > 0$.

05

Sum and Product of Slopes

$$ m_1 + m_2 = -\frac{2h}{b}, \quad m_1 m_2 = \frac{a}{b} $$

Rule: Derived from the auxiliary equation $bm^2 + 2hm + a = 0$; vital for problems involving angle bisectors or symmetric functions of slopes.

Application: Use this result in the stated type of problem.
06

Equation of Bisectors of Angles Between Lines

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

Rule: Direct joint equation of the internal and external bisectors of the angles between the lines $ax^2 + 2hxy + by^2 = 0$.

Application: Direct formula substitution questions in UP PGT Mathematics.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Assuming Real Lines When $h^2 < ab$

Common Pitfall: Candidates forget to check $h^2 \ge ab$ and calculate angles or slopes assuming lines are always real.

Counterexample: $x^2 + xy + y^2 = 0$, where $a=1, h=1/2, b=1 \implies h^2 - ab = 1/4 - 1 = -3/4 < 0$.

Why it fails: The slopes involve $\sqrt{-3}$, yielding complex imaginary lines. Calculating a real angle between them is algebraically invalid.

🛡️ Exam Defense: Always evaluate $h^2 - ab$ first before applying angle formulas.
⚠️ COMMON ERROR #02

Confusing $a+b=0$ with $h=0$

Common Pitfall: Mistaking $a+b=0$ (perpendicular lines) with $h=0$ (lines parallel to coordinate axes or coordinate axes themselves).

Counterexample: $x^2 - y^2 = 0$ ($a=1, h=0, b=-1$) vs $xy = 0$ ($a=0, h=1/2, b=0$).

Why it fails: $x^2 - y^2 = 0$ gives $y = \pm x$ (perpendicular, $a+b=0$), whereas $xy = 0$ gives coordinate axes ($h=1/2$).

🛡️ Exam Defense: Remember: $a+b=0$ governs orthogonality, while $h=0$ means the product term is absent.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Homogeneous Pair (Origin)$ax^2 + 2hxy + by^2 = 0$$bm^2 + 2hm + a = 0$Two lines intersecting at originNo linear or constant terms
Perpendicular Lines$a + b = 0$$a + b = 0$Angle between lines is $90^\circ$Sum of coefficients of $x^2$ and $y^2$ is zero
Coincident Lines$h^2 = ab$$h^2 = ab$Two identical overlapping linesDiscriminant of auxiliary equation is zero
Angle Bisectors$h(x^2 - y^2) = (a - b)xy$$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$Perpendicular lines bisecting original angleKeywords: 'bisectors of angles'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the value of $k$ for which the equation $2x^2 + 7xy + 3y^2 + 8x + 14y + k = 0$ represents a pair of straight lines.

(A) $8$(B) $6$(C) $10$(D) $12$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Identify general second degree equation coefficients $a=2, b=3, c=k, h=7/2, g=4, f=7$. Step 2: Apply condition for pair of lines: $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. Step 3: Substitute values: $2(3)(k) + 2(7)(4)(7/2) - 2(7)^2 - 3(4)^2 - k(7/2)^2 = 0$. Step 4: Solve for $k$ to get $k = 8$.
⚡ Method B: 25-Second Exam Shortcut
Use determinant expansion $\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \\ \end{vmatrix} = 0$ directly on the scratchpad for rapid calculation.
💡 Tip: This is a recurring 1-minute question in UP PGT. Memorize the determinant condition.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If the angle between the two lines represented by $ax^2 + 2hxy + by^2 = 0$ is $60^\circ$, then show that $(a+b)^2 = 4(h^2 - ab)$.

(A) $(a+b)^2 = 3(h^2 - ab)$(B) $(a+b)^2 = 4(h^2 - ab)$(C) $(3a+b)^2 = 4(h^2 - ab)$(D) $(a+3b)^2 = 4(h^2 - ab)$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Write angle formula $\tan 60^\circ = \frac{2\sqrt{h^2 - ab}}{a+b}$. Step 2: Substitute $\tan 60^\circ = \sqrt{3}$. Step 3: Square both sides: $3 = \frac{4(h^2 - ab)}{(a+b)^2}$. Step 4: Rearrange to get $(a+b)^2 = \frac{4}{3}(h^2 - ab)$, wait, checking options: $(a+b)^2 = 3(h^2 - ab)$ when squared properly ($3 = 4(h^2-ab)/(a+b)^2 \implies 3(a+b)^2 = 4(h^2-ab)$? Let's verify: $\tan^2 60 = 3 = \frac{4(h^2-ab)}{(a+b)^2} \implies 3(a+b)^2 = 4(h^2-ab)$).
⚡ Method B: 25-Second Exam Shortcut
Square the standard tangent formula immediately upon reading 'angle is $60^\circ$' ($3 = \frac{4(h^2-ab)}{(a+b)^2}$).
💡 Tip: Be extremely careful with squaring factors of 3 and 4 in exam pressure.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: The equation $x^2 - 5xy + 6y^2 = 0$ represents:
A A circle
B A pair of parallel lines
C A pair of real and distinct lines through origin
D A pair of imaginary lines
Derivation & Analysis:

Here $a=1, h=-5/2, b=6$. Discriminant $h^2 - ab = 25/4 - 6 = 1/4 > 0$. Thus, it represents two real and distinct lines through the origin.

Q2
MCQ 2: The acute angle between the pair of lines $x^2 - 7xy + 12y^2 = 0$ is:
A $0^\circ$
B $30^\circ$
C $45^\circ$
D $60^\circ$
Derivation & Analysis:

Factorizing $x^2 - 7xy + 12y^2 = (x - 3y)(x - 4y) = 0$. Slopes are $m_1 = 3, m_2 = 4$. Angle $\tan\theta = |\frac{3-4}{1 + 12}| = \frac{1}{13}$, or using formula $\tan\theta = \frac{2\sqrt{(-7/2)^2 - 12}}{1+12} = \frac{2\sqrt{49/4 - 48/4}}{13} = \frac{1}{13}$... wait, let's use standard values. If $a=1, h=-5/2, b=6$, slopes are $2, 3$. Let's check $x^2 - 5xy + 6y^2 = 0$, slopes are 2 and 3, $\tan\theta = \frac{3-2}{1+6} = 1/7$. For $x^2 - 7xy + 12y^2 = 0$, $\tan\theta = |\frac{3-4}{1+12}| = 1/13$.

Q3
MCQ 3: If the lines $ax^2 + 2hxy + by^2 = 0$ are coincident, then:
A $h^2 = ab$
B $h^2 > ab$
C $h^2 < ab$
D $a+b = 0$
Derivation & Analysis:

Coincident lines require the auxiliary quadratic $bm^2 + 2hm + a = 0$ to have equal roots, which means discriminant $4h^2 - 4ab = 0 \implies h^2 = ab$.

Q4
MCQ 4: If the slope of one line of the pair $ax^2 + 2hxy + by^2 = 0$ is thrice the slope of the other, then:
A $3h^2 = 8ab$
B $8h^2 = 3ab$
C $3h^2 = 16ab$
D $16h^2 = 3ab$
Derivation & Analysis:

Let roots be $m$ and $3m$. Sum $4m = -2h/b \implies m = -h/2b$. Product $3m^2 = a/b \implies 3(-h/2b)^2 = a/b \implies 3h^2/4b^2 = a/b \implies 3h^2 = 4ab$... wait, $3h^2 = 4ab \times b$ ? Let's re-verify: $3(h^2/4b^2) = a/b \implies 3h^2 = 4ab$. Thus $3h^2 = 4ab$.

Q5
MCQ 5: The combined equation of the lines passing through the origin and perpendicular to the lines $ax^2 + 2hxy + by^2 = 0$ is:
A $bx^2 - 2hxy + ay^2 = 0$
B $ax^2 - 2hxy + by^2 = 0$
C $bx^2 + 2hxy + ay^2 = 0$
D $ay^2 + 2hxy + bx^2 = 0$
Derivation & Analysis:

Replacing $x$ with $y$ and $y$ with $-x$ (or negating slopes) transforms $ax^2 + 2hxy + by^2 = 0$ into $a(y)^2 + 2h(y)(-x) + b(-x)^2 = 0 \implies by^2 - 2hxy + ax^2 = 0$, which rearranges to $bx^2 - 2hxy + ay^2 = 0$.

Q6
MCQ 6: 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:

The product of coefficients of $xy$ for mutual angle bisectors yields $pq = -1$.

Q7
MCQ 7: The area of the triangle formed by the line $lx + my + n = 0$ and the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by:
A $\frac{n^2 \sqrt{h^2 - ab}}{am^2 - 2hlm + bl^2}$
B $\frac{n^2 \sqrt{h^2 - ab}}{|am^2 - 2hlm + bl^2|}$
C $\frac{n \sqrt{h^2 - ab}}{am^2 - 2hlm + bl^2}$
D $\frac{n^2 (h^2 - ab)}{am^2 - 2hlm + bl^2}$
Derivation & Analysis:

Standard textbook formula for the area of the triangle formed by a line and homogeneous second-degree lines through the origin.

Q8
MCQ 8: Assertion (A): The equation $2x^2 + 3xy - 2y^2 = 0$ represents two perpendicular straight lines. Reason (R): For $ax^2 + 2hxy + by^2 = 0$ to represent perpendicular lines, $a + b = 0$.
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:

Here $a = 2, b = -2 \implies a + b = 2 + (-2) = 0$. Thus, (A) is true because of (R).

Q9
MCQ 9: If the lines represented by $3x^2 + 4xy - ky^2 = 0$ are perpendicular to each other, then the value of $k$ is:
A 3
B -3
C 1
D -1
Derivation & Analysis:

For perpendicular lines, $a + b = 0$. Here $a = 3$ and $b = -k$. Thus $3 + (-k) = 0 \implies k = 3$.

Q10
MCQ 10: The joint equation of the pair of lines passing through the origin and making an angle of $30^\circ$ with the line $y = x$ is:
A $x^2 - 4xy + y^2 = 0$
B $x^2 + 4xy + y^2 = 0$
C $3x^2 - 8xy + 3y^2 = 0$
D $3x^2 + 8xy + 3y^2 = 0$
Derivation & Analysis:

The line is $y = x$, angle with it is $30^\circ$. Using slope formula $\frac{m - 1}{1 + m} = \pm \tan 30^\circ = \pm \frac{1}{\sqrt{3}}$, solving gives slopes and subsequently the joint equation $3x^2 - 8xy + 3y^2 = 0$.

Common questions

Because substituting $x = 0$ and $y = 0$ satisfies the equation $ax^2 + 2hxy + by^2 = 0$ identically for any values of $a, h, b$. Furthermore, scaling any point $(x, y)$ by a factor $\lambda$ scales every term by $\lambda^2$, which cancels out, meaning the entire ray from the origin lies on the locus.

Candidates frequently confuse the condition for perpendicular lines ($a + b = 0$) with the condition for parallel lines or make sign errors when calculating the discriminant $h^2 - ab$ under exam time pressure.

Memorize the direct shortcuts: check $a+b=0$ instantly for orthogonality, use $\tan\theta = \frac{2\sqrt{h^2-ab}}{a+b}$ without re-deriving separate line equations, and utilize the auxiliary quadratic $bm^2 + 2hm + a = 0$ for slope relations.

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