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4. Coordinate Geometry of Two Dimensions

Classification as pair of straight lines

The general second-degree homogeneous and non-homogeneous equations in two variables represent a pair of straight lines under specific determinant conditions.

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

Condition for General Second Degree Equation to Represent a Pair of Straight Lines
The general equation of second degree in $x$ and $y$, given by $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$, represents a pair of straight lines if and only if the determinant of the coefficients vanishes, i.e., $$\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$$.

📐 Geometric Interpretation

Geometrically, the vanishing of the determinant ensures that the conic section degenerates from an ellipse, parabola, or hyperbola into two intersecting or parallel linear loci in the Cartesian plane.

⚙️ Algebraic & Structural Meaning

Algebraically, it implies that the quadratic polynomial in two variables can be factored into two linear factors with real or complex coefficients, corresponding to the left-hand side splitting into $(l_1 x + m_1 y + n_1)(l_2 x + m_2 y + n_2) = 0$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Vanishing of the 3x3 Determinant ($\Delta = 0$) Real algebraic constraint Strictly Mandatory / Necessary If $\Delta \neq 0$, the equation represents a non-degenerate conic (circle, ellipse, parabola, or hyperbola) rather than straight lines.
Condition for Real and Distinct Lines ($h^2 > ab$) Inequality constraint on quadratic coefficients Essential for real distinct intersection If $h^2 < ab$, the lines are imaginary with a real point of intersection; if $h^2 = ab$, the lines are coincident.
Condition for Parallel Lines ($af^2 = bg^2$ and $h^2 = ab$) Parameter matching Essential for parallel/coincident classification If not satisfied when lines are parallel, the equation does not represent valid parallel straight lines.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Determinant Zero Check for Second Degree Equation

$$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$$

Rule: Substitute coefficients directly into the expanded determinant formula to verify line pair condition.

Application: undefined
02

Angle Between the Pair of Lines

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

Rule: If $a + b = 0$, the lines are perpendicular (orthogonal). If $h^2 = ab$, the lines are coincident or parallel.

03

Point of Intersection of the Line Pair

$$x = \frac{hf - bg}{ab - h^2}, \quad y = \frac{gh - af}{ab - h^2}$$

Rule: Obtained by partially differentiating the expression with respect to $x$ and $y$ and solving the linear system $\frac{\partial S}{\partial x} = 0$ and $\frac{\partial S}{\partial y} = 0$.

Application: Saves massive calculation time compared to factorization in DSSSB and UP PGT.
04

Distance Between Two Parallel Lines

$$d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}} \quad \text{or} \quad 2\sqrt{\frac{f^2 - bc}{b(a + b)}}$$

Rule: Use this formula directly when $h^2 = ab$ and the lines are recognized as parallel.

Application: Asked repeatedly in UP TGT/PGT coordinate geometry sections.
05

Combined Equation of Lines Through Origin

$$ax^2 + 2hxy + by^2 = 0$$

Rule: Homogeneous second-degree equations always pass through the origin $(0,0)$. Sum of slopes $= -\frac{2h}{b}$, product of slopes $= \frac{a}{b}$.

06

Rapid Option Elimination via Perpendicularity

$$a + b = 0$$

Rule: Instantly check $a + b = 0$ to spot perpendicular line pairs without expanding options.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Assuming $\Delta = 0$ Guarantees Real Lines

Common Pitfall: Candidates check $\Delta = 0$ and conclude lines are real, ignoring the condition $h^2 \ge ab$.

Counterexample: $x^2 + y^2 + 1 = 0$ or similar degenerate forms with $h^2 < ab$

Why it fails: Even if $\Delta = 0$, if $h^2 < ab$, the represented lines have real point of intersection but imaginary component equations.

🛡️ Exam Defense: Always check $h^2 - ab$ sign to classify real vs imaginary lines.
⚠️ COMMON ERROR #02

Misidentifying $2h$ Coefficient

Common Pitfall: Forgetting to divide the coefficient of $xy$ by $2$ when extracting $h$.

Counterexample: Given $2xy$, $h$ is $1$, not $2$.

Why it fails: Leads to completely incorrect determinant calculation and wrong parameter values.

🛡️ Exam Defense: Always write the equation in standard form: $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ before identifying $a, b, c, f, g, h$.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Homogeneous Pair$c=0, g=0, f=0$$ax^2 + 2hxy + by^2 = 0$Pair of lines passing through originNo linear terms present
General Pair$\Delta = 0$$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$Pair of intersecting or parallel linesFull second degree with linear terms
Perpendicular Lines$a + b = 0$$x^2 + y^2 + \dots = 0$ with $a+b=0$Lines intersect at $90^\circ$Sum of coefficients of $x^2$ and $y^2$ is zero
Parallel Lines$h^2 = ab$ and $af^2 = bg^2$Distance formula appliesNon-intersecting parallel linesDiscriminant condition for parallel conics

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

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

(A) $6$(B) $8$(C) $10$(D) $12$
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Identify $a=2, b=3, h=7/2, g=4, f=7, c=\lambda$. Apply $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. Substitute values and solve for $\lambda = 8$.
⚡ Method B: 25-Second Exam Shortcut
Use the determinant expansion directly: $2(3\lambda - 49) + 7(28 - 4\lambda) + 4(49 - 12) = 0$, giving $\lambda = 8$ instantly.
💡 Tip: Be extremely careful with fractions when $h$ is odd.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If the equation $x^2 - 5xy + 6y^2 + x - 3y + k = 0$ represents a pair of straight lines, then find $k$.

(A) $0$(B) $1$(C) $2$(D) $-1$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Here $a=1, b=6, h=-5/2, g=1/2, f=-3/2, c=k$. Compute $\Delta = 0$ to get $k = 0$.
⚡ Method B: 25-Second Exam Shortcut
Factorize the quadratic part: $(x - 2y)(x - 3y) + x - 3y + k = 0 \implies (x - 2y + 1)(x - 3y + k') = 0$, equating constant terms gives $k = 0$.
💡 Tip: Factorization of homogeneous part works wonders for quick verification.

Practice questions

Choose an answer to check it and see the solution:

Q1
The equation $12x^2 + 7xy - 12y^2 - x + 7y - 1 = 0$ represents:
A A circle
B A pair of parallel straight lines
C A pair of perpendicular straight lines
D An ellipse
Derivation & Analysis:

Here $a = 12$ and $b = -12$. Since $a + b = 12 + (-12) = 0$, the lines are perpendicular.

Q2
The angle between the pair of straight lines given by $x^2 - 4xy + y^2 = 0$ is:
A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$
Derivation & Analysis:

Here $a=1, b=1, h=-2$. $\tan \theta = \frac{2\sqrt{h^2-ab}}{a+b} = \frac{2\sqrt{4-1}}{2} = \sqrt{3}$, so $\theta = \frac{\pi}{3}$.

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

For coincident lines, the angle between them is zero, which requires $\sqrt{h^2 - ab} = 0 \implies h^2 = ab$.

Q4
The point of intersection of the lines $2x^2 - 5xy + 2y^2 + 5x - 5y + 2 = 0$ is:
A (1, 1)
B (-1, -1)
C (0, 0)
D (2, 1)
Derivation & Analysis:

Solving $\frac{\partial S}{\partial x} = 4x - 5y + 5 = 0$ and $\frac{\partial S}{\partial y} = -5x + 4y - 5 = 0$ yields $x = -1, y = -1$.

Q5
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 $x$ by $y$ and $y$ by $-x$ gives the perpendicular line pair: $b(-y)^2 + 2h(-y)(x) + a(x)^2 = 0 \implies bx^2 - 2hxy + ay^2 = 0$.

Q6
If the slope of one line of the pair $ax^2 + 2hxy + by^2 = 0$ is $n$ times the slope of the other, then:
A $4nh^2 = ab(n+1)^2$
B $4nh^2 = ab(n-1)^2$
C $nh^2 = ab(n+1)^2$
D $h^2 = 4abn$
Derivation & Analysis:

Let slopes be $m_1$ and $m_2 = nm_1$. Sum $m_1 + nm_1 = -2h/b$, product $nm_1^2 = a/b$. Eliminating $m_1$ yields $4nh^2 = ab(n+1)^2$.

Q7
The value of $k$ for which $3x^2 + 10xy + 8y^2 - 14x - 22y + k = 0$ represents a pair of straight lines is:
A 6
B 8
C 10
D 15
Derivation & Analysis:

Substituting coefficients into $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ yields $k = 15$.

Q8
The distance between the parallel lines given by $4x^2 + 12xy + 9y^2 - 6x - 9y + 2 = 0$ is:
A $\frac{1}{\sqrt{13}}$
B $\frac{2}{\sqrt{13}}$
C $\frac{1}{13}$
D $0$
Derivation & Analysis:

Factoring the parallel lines or applying the distance formula $2\sqrt{\frac{g^2 - ac}{a(a+b)}}$ gives $\frac{2}{\sqrt{13}}$.

Q9
If the second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents two parallel straight lines, then:
A $h^2 = ab$ and $af^2 = bg^2$
B $h^2 > ab$ and $f^2 > gc$
C $h^2 < ab$
D $abc = fgh$
Derivation & Analysis:

Parallel lines require $h^2 = ab$ (to ensure parallel direction) and $af^2 = bg^2$ (consistency of linear terms).

Q10
The area of the triangle formed by the pair of lines $ax^2 + 2hxy + by^2 = 0$ and the line $lx + my + n = 0$ is:
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 $\frac{n \sqrt{h^2 - ab}}{am^2 + bl^2}$
D $\frac{n^2 (h^2 - ab)}{am^2 + bl^2}$
Derivation & Analysis:

Standard textbook formula for the area of the triangle formed by origin-vertex line pairs and a transversal line is $\frac{n^2 \sqrt{h^2 - ab}}{|am^2 - 2hlm + bl^2|}.

Common questions

When $\Delta = 0$, if $h^2 > ab$, the lines are real and intersecting. If $h^2 = ab$, the lines are parallel or coincident.

Failure to properly account for the $2h$, $2g$, and $2f$ coefficients in the standard equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is the primary reason for calculation errors.

Memorize the formula $\tan \theta = \pm \frac{2\sqrt{h^2 - ab}}{a + b}$ and immediately check if $a + b = 0$ to spot perpendicular lines without calculation.

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