Pair of Straight lines | NEB Class 11 Mathematics Notes

Chapter Overview

Namaste students! This chapter is a vital part of your NEB Class 11 Mathematics syllabus, expanding on your foundational knowledge of straight lines. Instead of analyzing one line at a time, we will learn how a single second-degree equation can represent two straight lines simultaneously. In NEB board exams, pair of lines questions are highly frequent, covering everything from 1-mark MCQs to 5-mark long algebraic proofs. Mastering this chapter provides the conceptual basis for Class 12 Conic Sections.

What this chapter covers

Combined Equation of Two LinesHomogeneous Equation of Second DegreeNature of Lines and Slope RelationshipsAngle Between a Pair of LinesConditions for Parallel, Coincident, and Perpendicular LinesEquation of Angle BisectorsGeneral Second-Degree Equation Representing a Pair of LinesHomogenization of Curves with Straight Lines

Learning Objectives

1

Form a combined second-degree equation from two separate straight line equations

2

Deconstruct a given second-degree equation into its two separate line equations

3

Calculate the angle between two lines represented by a homogeneous equation

4

Identify and apply the algebraic conditions for lines to be parallel, coincident, or perpendicular

5

Find the joint equation of the bisectors of angles between two lines

6

Use the general discriminant condition to determine if a second-degree equation represents a pair of straight lines

The Magic of Homogenization

Key Insight

The second-degree homogeneous equation ax² + 2hxy + by² = 0 always represents a pair of straight lines passing through the origin. When a curve and a straight line intersect, you can make the curve equation homogeneous of degree 2 using the line equation. This combined equation yields the equations of the lines joining the origin to those points of intersection.

In NEB exams: always use the relation $1 = \frac{lx+my}{n}$ to substitute into the constant and linear terms of a curve equation. This is the homogenization technique.

Key Definitions

Combined Equation

The single second-degree equation obtained by setting two individual straight line equations to zero and multiplying them together.

$$(L_1) \times (L_2) = 0$$

Homogeneous Equation of Second Degree

An algebraic equation where the sum of powers of the variables in every single term is exactly 2.

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

General Equation of Second Degree

The general algebraic expression of second degree containing linear, quadratic, and cross-product terms.

$$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$

Angle Bisectors

Two mutually perpendicular lines that divide the angles formed by a pair of intersecting lines into two equal parts.

Coincident vs Perpendicular Lines

The coefficients $a$, $h$, and $b$ from the homogeneous equation $ax^2 + 2hxy + by^2 = 0$ determine how the two lines are oriented relative to each other.

Coincident (Parallel) Lines

$$h^2 = ab$$

The two lines lie directly on top of each other

Both lines pass through the origin, making them coincident rather than strictly parallel

The angle between the lines is 0° (tan θ = 0)

The value under the square root of the angle formula, $\sqrt{h^2 - ab}$, is exactly zero

VS

Perpendicular Lines

$$a + b = 0$$

The two lines intersect at exactly 90°

The product of their slopes satisfies $m_1 \cdot m_2 = a/b = -1$, yielding $a + b = 0$

The coefficients of $x^2$ and $y^2$ are equal in magnitude but opposite in sign

The denominator in the angle formula becomes zero, making $\tan \theta$ undefined

Quick check: Are the coefficients of $x^2$ and $y^2$ opposite in sign and equal? → Perpendicular. Does $h^2 = ab$? → Coincident.

Complete Formula Sheet with Derivations

Combined Equation of Two Lines

$$(a_1x + b_1y + c_1)(a_2x + b_2y + c_2) = 0$$

Represents the locus of points lying on either of the two individual straight lines.

1Let the two lines be $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$.

2Any point $P(x_1, y_1)$ lying on $L_1$ satisfies $L_1 = 0$, which makes the product $L_1 \cdot L_2 = 0$.

3Any point lying on $L_2$ satisfies $L_2 = 0$, which also makes the product $L_1 \cdot L_2 = 0$.

4Hence, the combined relation is: $$(a_1x + b_1y + c_1)(a_2x + b_2y + c_2) = 0$$

Angle Between Two Lines

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

Use the positive value for the acute angle. If a + b = 0, the lines are perpendicular.

1Let the two lines represented by $ax^2 + 2hxy + by^2 = 0$ be $y = m_1x$ and $y = m_2x$.

2Their combined equation is $(y - m_1x)(y - m_2x) = y^2 - (m_1+m_2)xy + m_1m_2x^2 = 0$.

3Dividing by $b$ and comparing with the standard form, we get: $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.

4Calculate difference of slopes: $m_1 - m_2 = \sqrt{(m_1+m_2)^2 - 4m_1m_2} = \sqrt{\frac{4h^2}{b^2} - \frac{4a}{b}} = \frac{2\sqrt{h^2-ab}}{b}$.

5Substitute into the angle formula $\tan \theta = \pm\frac{m_1 - m_2}{1 + m_1m_2}$:

6$$\tan \theta = \pm \frac{\frac{2\sqrt{h^2-ab}}{b}}{1 + \frac{a}{b}} = \pm \frac{2\sqrt{h^2 - ab}}{a + b}$$

Equation of Angle Bisectors

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

Represents a pair of mutually perpendicular lines bisecting the angles between the original lines.

1Let the original lines be $y - m_1x = 0$ and $y - m_2x = 0$.

2The equations of their individual bisectors are: $\frac{y - m_1x}{\sqrt{1+m_1^2}} = \pm \frac{y - m_2x}{\sqrt{1+m_2^2}}$.

3Squaring both sides and simplifying yields the joint equation of bisectors: $(m_1+m_2)(x^2 - y^2) + 2(1 - m_1m_2)xy = 0$.

4Substitute the slope relations $m_1+m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$ into this equation:

5$$-\frac{2h}{b}(x^2 - y^2) + 2\left(1 - \frac{a}{b}\right)xy = 0 \quad\Rightarrow\quad -h(x^2 - y^2) + (b-a)xy = 0$$

6Simplifying and rearranging terms gives the standard form: $$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$$

General Second-Degree Equation Condition

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

This is the condition for a non-homogeneous second-degree equation to represent a pair of lines.

1The general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ can be written as a quadratic in $x$:

2$$ax^2 + 2(hy+g)x + (by^2+2fy+c) = 0$$

3Solving for $x$ using the quadratic formula: $x = \frac{-2(hy+g) \pm \sqrt{4(hy+g)^2 - 4a(by^2+2fy+c)}}{2a}$.

4For this to represent two straight lines, the expression under the square root must be a perfect square in terms of $y$.

5This require the discriminant of the quadratic inside the square root to equal zero.

6Expanding and setting the discriminant to zero yields: $$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$$

Pair of Lines Decision Tree

Use this decision tree to identify the correct method or equation to solve pair of lines problems.

Are separate equations of two lines given?

Yes

Multiply the equations directly: (L₁) × (L₂) = 0 to find the combined equation.

No

Continue ↓

Do you need to find separate lines from ax² + 2hxy + by² = 0?

Yes

Factorize by splitting the middle term, or use quadratic formula for y in terms of x.

No

Continue ↓

Are you asked to find the angle between homogeneous lines?

Yes

Identify coefficients a, h, b and use: tan θ = ±2√(h²−ab) / (a+b).

No

Use homogenization (1 = (lx + my)/n) if finding lines joining the origin to curve intersections.

Slope Relations in Homogeneous Equations

The homogeneous equation of the second degree ax² + 2hxy + by² = 0 represents two straight lines passing through the origin. Their slopes are deeply connected to the coefficients.

$$m_1 + m_2 = -\frac{2h}{b} \quad\text{and}\quad m_1 \cdot m_2 = \frac{a}{b}$$

Why it works (Derivation):

1Let the two lines be $y = m_1x$ and $y = m_2x$, which can be written as $y - m_1x = 0$ and $y - m_2x = 0$.

2The joint equation is $(y - m_1x)(y - m_2x) = m_1m_2x^2 - (m_1+m_2)xy + y^2 = 0$.

3Comparing with the divided form $\frac{a}{b}x^2 + \frac{2h}{b}xy + y^2 = 0$ yields the slope relations: $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.

Example: For $x^2 - 6xy + 9y^2 = 0$, $a=1$, $2h=-6 \Rightarrow h=-3$, and $b=9$. Sum of slopes $m_1+m_2 = -(-6)/9 = 2/3$. Product of slopes $m_1 \cdot m_2 = 1/9$.

When to use: Use the sum and product of slopes whenever you encounter conditions relating individual slopes (e.g., 'one slope is twice the other' or 'slopes are in ratio 1:3').

Memory Tricks — Never Forget These

Trick: Four simple memory hooks to avoid common mathematical errors in this chapter.

Steps to Remember:

Halving the Middle: Always divide the $xy$ coefficient by 2 to find $h$. If the middle term is $8xy$, then $h = 4$, not $8$.
A plus B is Zero (Perpendicularity): If the sum of the coefficients of $x^2$ and $y^2$ is zero ($a+b=0$), the lines are perpendicular.
H-squared is AB (Coincidence): $h^2 = ab$ means coincident. Think 'Harmonious lines coincide'.
Bisector Rhythm: Memorize the joint bisector equation: 'x-squared minus y-squared over a minus b equals xy over h'.

Mnemonic: Halve the middle for h | a+b=0 for perp | h²=ab for coincident | Bisector: x²-y² over a-b equals xy over h

Common Mistakes

These typical errors cause significant mark losses in the NEB exams.

Using h instead of 2h: Students often take $h$ as the entire coefficient of $xy$. Always remember: the coefficient is $2h$, so divide it by 2.
Sign errors in a + b: When $b$ is negative, remember to preserve the sign. For $3x^2 + 5xy - 3y^2 = 0$, $a+b = 3 + (-3) = 0$ (perpendicular).
Homogenization constant confusion: When homogenizing, make sure the line equation is set to $1 = \frac{lx+my}{n}$. Never leave a constant on the right side of the curve equation non-homogenized.
Assuming parallel lines have separate origins: For homogeneous second-degree equations, parallel lines must coincide because both lines pass through the origin $(0,0)$.

NEB Exam Traps — Edge Cases

NEB Exam Warning

Examiners design questions specifically to test these conceptual traps:

The $x^2 + y^2 = 0$ trap: Students often say this represents a pair of lines. It only represents a single real point $(0,0)$, as the sum of two squares cannot be zero for any other real numbers.
Finding perpendicular lines through a point: To find lines passing through $(\alpha, \beta)$ perpendicular to $ax^2 + 2hxy + by^2 = 0$, use the formula $b(x-\alpha)^2 - 2h(x-\alpha)(y-\beta) + a(y-\beta)^2 = 0$. Notice that $a$ and $b$ swap places and the middle term changes its sign.
Distance between parallel lines: The formula $d = 2\sqrt{\frac{g^2-ac}{a(a+b)}}$ is only valid if $a+b > 0$ and the lines are parallel. Otherwise, factorize first, then use the basic parallel distance formula.

Question Recognition Guide

Keyword / Phrase in Question	Use This
Combine two separate lines $L_1 = 0$ and $L_2 = 0$	Multiply the equations: $L_1 \cdot L_2 = 0$
Determine separate equations from $ax^2 + 2hxy + by^2 = 0$	Factorize by splitting the middle term, or solve quadratic in x/y
Calculate the angle between homogeneous pair	Use: $\tan\theta = \pm\frac{2\sqrt{h^2 - ab}}{a+b}$
Find value of a parameter (like k) for representing a pair of lines	Apply: $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$
Equations of angle bisectors of $ax^2+2hxy+by^2=0$	Use: $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$
Find lines joining origin to intersection of line and curve	Homogenize the curve's equation using the relation: $1 = \frac{lx+my}{n}$
Check if lines are perpendicular	Check if $a+b = 0$
Check if lines are parallel or coincident	Check if $h^2 = ab$

Solved Examples (Easy → Hard)

1

Find the combined equation of $x - y = 0$ and $x + 2y = 0$.

Easy

Solution

Step 1 — Set up combination:
Multiply both equations: $(x - y)(x + 2y) = 0$.

Step 2 — Distribute and expand:
$x(x + 2y) - y(x + 2y) = 0$
$x^2 + 2xy - xy - 2y^2 = 0$
$$x^2 + xy - 2y^2 = 0$$

∴ x² + xy − 2y² = 0

2

Find separate equations for $x^2 - 7xy + 10y^2 = 0$.

Medium

Solution

Step 1 — Factorize by splitting the middle term:
$x^2 - 5xy - 2xy + 10y^2 = 0$.

Step 2 — Factor by grouping:
$x(x - 5y) - 2y(x - 5y) = 0$
$(x - 5y)(x - 2y) = 0$

Step 3 — Write separate equations:
Set each factor to zero:
$$x - 5y = 0 \quad \text{and} \quad x - 2y = 0$$

∴ x − 5y = 0 and x − 2y = 0

3

Find $k$ if $2x^2 + 3xy - 2y^2 + 3x + y + k = 0$ represents a pair of lines.

Hard

Solution

Step 1 — Compare with general form:
$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$
Here: $a=2, 2h=3 \Rightarrow h=1.5, b=-2, 2g=3 \Rightarrow g=1.5, 2f=1 \Rightarrow f=0.5, c=k$.

Step 2 — Apply pair of lines condition:
$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$

Substitute the values:
$(2)(-2)(k) + 2(0.5)(1.5)(1.5) - 2(0.5)^2 - (-2)(1.5)^2 - k(1.5)^2 = 0$
$-4k + 2.25 - 0.5 + 4.5 - 2.25k = 0$
$-6.25k + 6.25 = 0 \Rightarrow k = 1$.

∴ k = 1

MCQ Practice

Mixed Level — MCQ

1

The angle between lines $x^2 - y^2 = 0$ is:

Correct!

Incorrect. Correct: 90°

Solution

Comparing $x^2 - y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$, we get $a = 1$ and $b = -1$.
Since $a+b = 1 + (-1) = 0$, the condition for perpendicularity is met. Therefore, the angle between the lines is $90°$.

∴ 90°

2

If $h^2 = ab$, the lines represented by homogeneous equation are:

Correct!

Incorrect. Correct: Coincident

Solution

The condition for two lines represented by $ax^2 + 2hxy + by^2 = 0$ to be parallel or coincident is $h^2 = ab$.

∴ Coincident

3

For $x^2 + 4xy + ky^2 = 0$ to be coincident, the value of $k$ is:

Correct!

Incorrect. Correct: 4

Solution

Here $a = 1, 2h = 4 \Rightarrow h = 2, b = k$.
For coincident lines, $h^2 = ab \Rightarrow 2^2 = (1)(k) \Rightarrow k = 4$.

∴ 4

4

Sum of slopes of $x^2 - 6xy + y^2 = 0$ is:

Correct!

Incorrect. Correct: 6

Solution

Here $a=1, 2h=-6, b=1$.
Sum of slopes $m_1 + m_2 = -\frac{2h}{b} = -\frac{-6}{1} = 6$.

∴ 6

5

Product of slopes of $2x^2 + 5xy + 3y^2 = 0$ is:

Correct!

Incorrect. Correct: 2/3

Solution

Comparing coefficients: $a=2, b=3$.
Product of slopes $m_1 \cdot m_2 = \frac{a}{b} = \frac{2}{3}$.

∴ 2/3

6

The lines represented by $ax^2 + 2hxy + by^2 = 0$ always pass through:

Correct!

Incorrect. Correct: (0,0)

Solution

Since there are no constant or first-degree terms in the homogeneous equation, substituting $(x,y) = (0,0)$ always satisfies the equation. Thus, the lines always pass through the origin.

∴ (0,0)

7

If $a+b=0$, the angle between the lines is:

Correct!

Incorrect. Correct: π/2

Solution

Since $a+b = 0$ makes the denominator of $\tan\theta = \frac{2\sqrt{h^2-ab}}{a+b}$ zero, $\tan\theta = \infty \Rightarrow \theta = 90^\circ$ or $\pi/2$.

∴ π/2

8

Equation of bisectors of $x^2 - y^2 = 0$ is:

Correct!

Incorrect. Correct: xy = 0

Solution

Here $a=1, b=-1, h=0$.
Using $\frac{x^2-y^2}{a-b} = \frac{xy}{h} \Rightarrow \frac{x^2-y^2}{2} = \frac{xy}{0}$.
Cross-multiplying yields $xy = 0$.

∴ xy = 0

9

If $h^2 < ab$, the lines are:

Correct!

Incorrect. Correct: Imaginary

Solution

If $h^2 < ab$, then $h^2-ab < 0$, which makes the square root in the separate line slopes imaginary. Hence, the lines are imaginary but have a real point of intersection at the origin.

∴ Imaginary

10

The combined equation of $x=0$ and $y=0$ is:

Correct!

Incorrect. Correct: xy = 0

Solution

The combined equation of two lines $L_1 = 0$ and $L_2 = 0$ is $L_1 \cdot L_2 = 0$. Thus, $x \cdot y = 0 \Rightarrow xy = 0$.

∴ xy = 0

Short Questions (2 Marks Each)

1

Find separate equations of $2x^2 - 5xy + 2y^2 = 0$.

[2 marks]

Solution

Factorize by splitting the middle term:
$2x^2 - 4xy - xy + 2y^2 = 0$
$2x(x - 2y) - y(x - 2y) = 0$
$(2x - y)(x - 2y) = 0$
The separate lines are $2x - y = 0$ and $x - 2y = 0$.

∴ 2x - y = 0 and x - 2y = 0

2

Find the angle between $x^2 + 6xy + 9y^2 = 0$.

[2 marks]

Solution

Here $a = 1, 2h = 6 \Rightarrow h = 3, b = 9$.
Check condition: $h^2 - ab = 3^2 - (1)(9) = 9 - 9 = 0$.
Since $h^2 = ab$, the lines are coincident, so the angle is $0^\circ$.

∴ θ = 0°

3

Determine $k$ if $x^2 + kxy + y^2 = 0$ are perpendicular.

[2 marks]

Solution

The condition for perpendicularity is $a + b = 0$.
Here $a = 1$ and $b = 1$.
Since $a+b = 2 \neq 0$ regardless of $k$, this equation can never represent perpendicular lines for any real value of $k$.

∴ No real value of k

4

Find equation of bisectors for $x^2 + 4xy + y^2 = 0$.

[2 marks]

Solution

Here $a=1, b=1, h=2$.
Using bisector formula:
$$\frac{x^2 - y^2}{a - b} = \frac{xy}{h} \Rightarrow \frac{x^2 - y^2}{1 - 1} = \frac{xy}{2} \Rightarrow \frac{x^2 - y^2}{0} = \frac{xy}{2}$$
Cross-multiplying gives:
$$x^2 - y^2 = 0$$

∴ x² − y² = 0

5

Find combined equation of $x-2=0$ and $x-5=0$.

[2 marks]

Solution

Multiply both equations:
$(x - 2)(x - 5) = 0$
$$x^2 - 7x + 10 = 0$$

∴ x² − 7x + 10 = 0

6

Show lines $3x^2 + 4xy - 3y^2 = 0$ are perpendicular.

[2 marks]

Solution

Here $a = 3$ and $b = -3$.
Checking perpendicularity condition:
$a + b = 3 + (-3) = 0$.
Since $a+b=0$, the lines are perpendicular.

∴ Lines are perpendicular — proved

7

Find product of slopes of $x^2 - 10xy + 9y^2 = 0$.

[2 marks]

Solution

Here $a = 1$ and $b = 9$.
The product of slopes is given by:
$$m_1 \cdot m_2 = \frac{a}{b} = \frac{1}{9}$$

∴ Product of slopes = 1/9

8

Find separate lines for $(x+y)^2 - 4 = 0$.

[2 marks]

Solution

Rewrite using difference of squares $A^2 - B^2 = 0$:
$(x+y)^2 - 2^2 = 0$
$(x+y-2)(x+y+2) = 0$
The separate lines are $x+y-2=0$ and $x+y+2=0$.

∴ x + y − 2 = 0 and x + y + 2 = 0

9

Find angle between $x^2 - 3xy + 2y^2 = 0$.

[2 marks]

Solution

Here $a = 1, 2h = -3 \Rightarrow h = -1.5, b = 2$.
Using angle formula:
$$\tan\theta = \pm\frac{2\sqrt{h^2 - ab}}{a+b} = \pm\frac{2\sqrt{(-1.5)^2 - 1(2)}}{1+2} = \pm\frac{2\sqrt{2.25-2}}{3} = \pm\frac{2(0.5)}{3} = \pm\frac{1}{3}$$
Taking the acute angle: $\theta = \tan^{-1}(1/3)$.

∴ θ = tan⁻¹(1/3)

10

Find $k$ if $x^2 + 2xy + y^2 + k = 0$ passes through $(1,1)$.

[2 marks]

Solution

Substitute $(x,y) = (1,1)$ into the equation:
$(1)^2 + 2(1)(1) + (1)^2 + k = 0$
$1 + 2 + 1 + k = 0 \Rightarrow 4 + k = 0 \Rightarrow k = -4$.

∴ k = −4

Long Questions (4–5 Marks Each)

1

Prove that $x^2 - 5xy + 4y^2 + x + 2y - 2 = 0$ represents a pair of lines. Find their intersection.

[5 marks]

Solution

Step 1 — Identify coefficients:
$a=1, h=-2.5, b=4, g=0.5, f=1, c=-2$.

Step 2 — Verify pair of lines condition:
$abc + 2fgh - af^2 - bg^2 - ch^2$
$= (1)(4)(-2) + 2(1)(0.5)(-2.5) - 1(1)^2 - 4(0.5)^2 - (-2)(-2.5)^2$
$= -8 - 2.5 - 1 - 1 + 12.5 = 0$.
Since it equals 0, the equation represents a pair of straight lines.

Step 3 — Find intersection point:
Differentiate partially with respect to x: $2x - 5y + 1 = 0$ ... (i)
Differentiate partially with respect to y: $-5x + 8y + 2 = 0$ ... (ii)
Solving (i) and (ii):
From (i): $2x = 5y - 1 \Rightarrow x = 2.5y - 0.5$.
Substitute into (ii): $-5(2.5y - 0.5) + 8y + 2 = 0 \Rightarrow -12.5y + 2.5 + 8y + 2 = 0 \Rightarrow -4.5y + 4.5 = 0 \Rightarrow y=1$.
Thus $x = 2.5(1) - 0.5 = 2$.
The point of intersection is $(2, 1)$.

∴ Intersection point = (2, 1) — proved

2

Find the equation of lines joining origin to intersection of $x+y=1$ and $x^2+y^2=4$.

[4 marks]

Solution

Step 1 — Express 1 from the line equation:
From $x+y=1$, we have $1 = x+y$.

Step 2 — Homogenize the curve equation:
Multiply constant terms to make all terms second degree:
$x^2 + y^2 = 4(1)^2 \Rightarrow x^2 + y^2 = 4(x+y)^2$.

Step 3 — Expand and simplify:
$x^2 + y^2 = 4(x^2 + 2xy + y^2)$
$x^2 + y^2 = 4x^2 + 8xy + 4y^2$
$$3x^2 + 8xy + 3y^2 = 0$$

∴ 3x² + 8xy + 3y² = 0

3

Show that bisectors of $ax^2 + 2hxy + by^2 = 0$ are always perpendicular.

[4 marks]

Solution

Step 1 — Write the combined equation of bisectors:
$$\frac{x^2 - y^2}{a - b} = \frac{xy}{h} \Rightarrow h(x^2 - y^2) = (a-b)xy \Rightarrow hx^2 - (a-b)xy - hy^2 = 0$$

Step 2 — Check the condition for perpendicularity:
Compare with standard form $Ax^2 + 2Hxy + By^2 = 0$.
Here: $A = h$ and $B = -h$.
Sum of coefficients: $A + B = h + (-h) = 0$.
Since the sum of the coefficients of $x^2$ and $y^2$ is zero, the angle bisectors are always perpendicular to each other.

∴ A + B = 0 — proved

4

Find the area of triangle formed by $y^2 - 5xy + 6x^2 = 0$ and $y=4$.

[5 marks]

Solution

Step 1 — Find separate lines for $6x^2 - 5xy + y^2 = 0$:
$(3x - y)(2x - y) = 0 \Rightarrow y = 2x \text{ and } y = 3x$.

Step 2 — Find vertices of the triangle:
- Intersection of $y=2x$ and $y=3x$ is the origin $O(0,0)$.
- Intersection of $y=2x$ and $y=4$ is $A(2,4)$.
- Intersection of $y=3x$ and $y=4$ is $B(4/3, 4)$.

Step 3 — Calculate area:
Using area formula with vertices $(0,0)$, $(2,4)$, and $(4/3, 4)$:
$$\text{Area} = \frac{1}{2}|0(4-4) + 2(4-0) + \frac{4}{3}(0-4)| = \frac{1}{2}|8 - \frac{16}{3}| = \frac{1}{2}|\frac{8}{3}| = \frac{4}{3} \text{ sq. units}$$

∴ Area = 4/3 square units

5

If the lines joining origin to intersection of $y=mx+c$ and $x^2+y^2=a^2$ are perpendicular, prove $2c^2 = a^2(1+m^2)$.

[5 marks]

Solution

Step 1 — Get homogenization term:
From $y - mx = c \Rightarrow \frac{y - mx}{c} = 1$.

Step 2 — Homogenize $x^2 + y^2 = a^2$:
$x^2 + y^2 = a^2\left(\frac{y - mx}{c}\right)^2$
$c^2(x^2 + y^2) = a^2(y^2 - 2mxy + m^2x^2)$
$(c^2 - a^2m^2)x^2 + 2a^2mxy + (c^2 - a^2)y^2 = 0$.

Step 3 — Apply perpendicular condition:
Sum of coefficients of $x^2$ and $y^2$ must be 0:
$(c^2 - a^2m^2) + (c^2 - a^2) = 0$
$2c^2 - a^2(m^2 + 1) = 0 \Rightarrow 2c^2 = a^2(1+m^2)$. Hence proved.

∴ 2c² = a²(1 + m²) — proved

6

Separate $x^2 + 2xy\sec\theta + y^2 = 0$.

[4 marks]

Solution

Step 1 — Treat as a quadratic in x:
$x^2 + (2y\sec\theta)x + y^2 = 0$.

Step 2 — Apply quadratic formula:
$x = \frac{-2y\sec\theta \pm \sqrt{(2y\sec\theta)^2 - 4(1)(y^2)}}{2}$
$x = \frac{-2y\sec\theta \pm \sqrt{4y^2\sec^2\theta - 4y^2}}{2} = \frac{-2y\sec\theta \pm 2y\sqrt{\sec^2\theta - 1}}{2}$
$x = -y\sec\theta \pm y\tan\theta = -y(\sec\theta \mp \tan\theta)$.

Step 3 — Separate lines:
$x + (\sec\theta - \tan\theta)y = 0$ and $x + (\sec\theta + \tan\theta)y = 0$.

∴ x + (secθ − tanθ)y = 0 and x + (secθ + tanθ)y = 0

7

Find distance between parallel lines $x^2 + 4xy + 4y^2 - 6x - 12y + 8 = 0$.

[5 marks]

Solution

Step 1 — Factorize the second-degree terms:
$x^2 + 4xy + 4y^2 = (x+2y)^2$.

Step 2 — Rewrite the whole equation:
$(x+2y)^2 - 6(x+2y) + 8 = 0$.
Let $u = x+2y$: $u^2 - 6u + 8 = 0 \Rightarrow (u-2)(u-4) = 0$.
So separate lines are $x+2y-2=0$ and $x+2y-4=0$.

Step 3 — Find distance:
$$d = \frac{|c_1 - c_2|}{\sqrt{a^2+b^2}} = \frac{|-2 - (-4)|}{\sqrt{1^2+2^2}} = \frac{2}{\sqrt{5}}$$ units.

∴ d = 2/√5 units

Past NEB Board Questions

1

Find the equation of the angle bisectors of the lines $3x-4y+7=0$ and $12x+5y-2=0$. Also find which bisector passes through the origin.

NEB 2078 [5 marks]

Solution

Step 1 — Setup bisector equation:
$$\frac{3x-4y+7}{\sqrt{3^2+(-4)^2}} = \pm\frac{12x+5y-2}{\sqrt{12^2+5^2}} \Rightarrow \frac{3x-4y+7}{5} = \pm\frac{12x+5y-2}{13}$$

Step 2 — Cross multiply:
(+) Case: $13(3x-4y+7) = 5(12x+5y-2) \Rightarrow 21x + 77y - 101 = 0$
(−) Case: $13(3x-4y+7) = -5(12x+5y-2) \Rightarrow 99x - 27y + 81 = 0 \Rightarrow 11x - 3y + 9 = 0$

Step 3 — Find bisector through origin:
Make the constant terms positive. $3x-4y+7=0$ ($c_1>0$) and $-12x-5y+2=0$ ($c_2>0$).
The positive case contains the origin: $13(3x-4y+7) = 5(-12x-5y+2) \Rightarrow 11x - 3y + 9 = 0$.

∴ Bisectors: 21x + 77y − 101 = 0 and 11x − 3y + 9 = 0. Bisector containing origin: 11x − 3y + 9 = 0.

2

Find the value of $k$ if $2x^2 + 7xy + 3y^2 - 4x - 7y + k = 0$ is a pair of lines.

NEB 2079 [4 marks]

Solution

Step 1 — Identify coefficients:
$a=2, 2h=7 \Rightarrow h=3.5, b=3, 2g=-4 \Rightarrow g=-2, 2f=-7 \Rightarrow f=-3.5, c=k$.

Step 2 — Apply pair of lines condition:
$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$
$(2)(3)(k) + 2(-3.5)(-2)(3.5) - 2(-3.5)^2 - 3(-2)^2 - k(3.5)^2 = 0$
$6k + 49 - 24.5 - 12 - 12.25k = 0$
$-6.25k + 12.5 = 0 \Rightarrow k = 2$.

∴ k = 2

Chapter Test — Full Mixed Paper

Attempt this test in 45 minutes to benchmark your understanding of pair of lines.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

The angle between the lines represented by $xy = 0$ is:

Correct!

Incorrect. Correct: 90°

Solution

The lines are $x = 0$ and $y = 0$, which are the coordinate axes. The angle between them is $90^\circ$.

∴ 90°

2

If $h^2 - ab > 0$, the lines are:

Correct!

Incorrect. Correct: Real and distinct

Solution

When $h^2 - ab > 0$, the square root yields a real positive number, giving two distinct real values for slopes.

∴ Real and distinct

2 Section B — Short Questions (2 Marks Each)

1

Separate the lines represented by $x^2 - 4y^2 = 0$.

[2 marks]

Solution

Using $a^2 - b^2 = (a-b)(a+b)$:
$(x - 2y)(x + 2y) = 0$.
So separate lines are $x - 2y = 0$ and $x + 2y = 0$.

∴ x − 2y = 0 and x + 2y = 0

2

Find the angle between the lines $2x^2 + xy - y^2 = 0$.

[2 marks]

Solution

Identify coefficients: $a=2, h=0.5, b=-1$.
$$\tan\theta = \pm\frac{2\sqrt{(0.5)^2 - 2(-1)}}{2 + (-1)} = \pm\frac{2\sqrt{0.25 + 2}}{1} = \pm2\sqrt{2.25} = \pm2(1.5) = \pm3$$
Acute angle $\theta = \tan^{-1}(3)$.

∴ θ = tan⁻¹(3)

3 Section C — Long Questions (4–5 Marks Each)

1

Find the joint equation of bisectors of angles between the lines $x^2 - 3xy + 2y^2 = 0$.

[4 marks]

Solution

Identify coefficients: $a=1, 2h=-3 \Rightarrow h=-1.5, b=2$.
Apply formula: $\frac{x^2 - y^2}{a-b} = \frac{xy}{h}$
$$\frac{x^2 - y^2}{1-2} = \frac{xy}{-1.5} \Rightarrow \frac{x^2 - y^2}{-1} = \frac{xy}{-1.5}$$
$$1.5(x^2 - y^2) = xy \Rightarrow 3(x^2 - y^2) = 2xy \Rightarrow 3x^2 - 2xy - 3y^2 = 0$$

∴ 3x² − 2xy − 3y² = 0

One-Page Revision Cheat Sheet

1

Origin Lines: Any homogeneous second-degree equation $ax^2 + 2hxy + by^2 = 0$ represents two straight lines passing through the origin.

2

Slopes: Sum of slopes is $m_1+m_2 = -2h/b$. Product of slopes is $m_1 \cdot m_2 = a/b$. Always identify $h$ correctly by halving the $xy$ coefficient.

3

Angle: Calculated using $\tan\theta = \left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$. Use positive sign for acute, negative for obtuse.

4

Perpendicular: If $a+b=0$ (the sum of coefficients of $x^2$ and $y^2$ is zero), the lines are perpendicular.

5

Coincident: If $h^2 = ab$, the two lines are coincident (lie on top of each other).

6

Bisector Joint Equation: Always given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$. Sum of coefficients is $h + (-h) = 0$, meaning bisectors are always perpendicular.

7

General Line Pair: $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents straight lines if and only if $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.

8

Homogenization: To connect origin with intersections of curve and line $lx+my=n$, rewrite line as $1 = \frac{lx+my}{n}$ and substitute to make curve terms homogeneous of degree 2.