IIT JEE Main Maths -Unit 10- Pair of Straight Line- Study Notes-New Syllabus

IIT JEE Main Maths -Unit 10- Pair of Straight Line – New syllabus

IIT JEE Main Maths -Unit 10- Pair of Straight Line -IIT JEE Main Maths – per latest Syllabus.

Key Concepts:

  • Family of Lines

IIT JEE Main Maths -Study Notes – All Topics

Family of Lines

A family of lines is a group of infinitely many straight lines that satisfy a common condition. In coordinate geometry, families of lines help in finding equations of lines that pass through specific sets of points or satisfy geometric conditions.

Family of Lines Passing Through the Intersection of Two Lines

If two lines are:

\( L_1 \equiv a_1x + b_1y + c_1 = 0 \)
\( L_2 \equiv a_2x + b_2y + c_2 = 0 \)

Then every line passing through their intersection is given by:

\( L_1 + \lambda L_2 = 0 \)

This is the most important formula in JEE.

Expanding:

\( (a_1 + \lambda a_2)x + (b_1 + \lambda b_2)y + (c_1 + \lambda c_2) = 0 \)

 Family of Lines Through a Given Point \( (x_1, y_1) \)

The general form is:

\( (y – y_1) = m(x – x_1) \)

where parameter \( m \) varies over all real numbers.

This represents the family of all lines passing through a fixed point.

 Family of Lines Parallel to a Given Line

If line has slope \( m \), then every line parallel to it is of the form:

\( y = mx + c \)

where \( c \) is any constant.

Thus parallel family depends on the constant \( c \).

Family of Lines Perpendicular to a Given Line

If slope of given line = \( m \), then every line perpendicular has slope:

\( -\dfrac{1}{m} \)

Thus general perpendicular family:

\( y = -\dfrac{1}{m}x + c \)

 Combined Family (Most Useful in JEE)

If you want a line that:

  • passes through intersection of two given lines
  • and satisfies some additional condition

Then use

\( L_1 + \lambda L_2 = 0 \)

and find \( \lambda \) using the condition.

 Special Results

  • If the required line must pass through a specific point → substitute the point and solve for \( \lambda \).
  • If the required line must be perpendicular to another line → use slope condition to find \( \lambda \).
  • If the required line must be at a fixed distance from origin → use distance formula.
  • If the required line must form a given angle → use angle-of-lines formula.

Example 

Find the family of lines passing through the intersection of \( x + y – 2 = 0 \) and \( 2x – y + 3 = 0 \).

▶️ Answer / Explanation

General family:

\( (x + y – 2) + \lambda(2x – y + 3) = 0 \)

Expand:

\( x + y – 2 + \lambda(2x – y + 3) = 0 \)

This is the required family.

Example 

Find the equation of the line passing through the intersection of \( 3x + y – 5 = 0 \) and \( x – 2y + 4 = 0 \), and also passing through the point \( (1, 2) \).

▶️ Answer / Explanation

Family: \( (3x + y – 5) + \lambda(x – 2y + 4) = 0 \)

Substitute point \( (1, 2) \):

\( 3(1) + 2 – 5 + \lambda(1 – 4 + 4) = 0 \)

\( 0 + \lambda(1) = 0 \Rightarrow \lambda = 0 \)

Thus required line:

\( 3x + y – 5 = 0 \)

Example 

Find the line passing through the intersection of \( x – y + 1 = 0 \) and \( 2x + 3y – 5 = 0 \), and perpendicular to the line \( 3x – y + 7 = 0 \).

▶️ Answer / Explanation

Step 1: Form the family

\( (x – y + 1) + \lambda(2x + 3y – 5) = 0 \)

Equation becomes:

\( (1 + 2\lambda)x + (-1 + 3\lambda)y + (1 – 5\lambda) = 0 \)

Step 2: Use perpendicularity condition

Line perpendicular to \( 3x – y + 7 = 0 \) has slope = negative reciprocal.

Slope of given line = \( m = \dfrac{3}{1} = 3 \)

Required slope = \( -\dfrac{1}{3} \)

Slope of family line:

\( m_f = -\dfrac{1 + 2\lambda}{-1 + 3\lambda} \)

Set equal:

\( -\dfrac{1 + 2\lambda}{-1 + 3\lambda} = -\dfrac{1}{3} \)

Cross multiply:

\( 3(1 + 2\lambda) = 1(-1 + 3\lambda) \)

\( 3 + 6\lambda = -1 + 3\lambda \)

\( 3\lambda = -4 \Rightarrow \lambda = -\dfrac{4}{3} \)

Step 3: Substitute back

\( 1 + 2\lambda = 1 – \dfrac{8}{3} = -\dfrac{5}{3} \)

\( -1 + 3\lambda = -1 – 4 = -5 \)

\( 1 – 5\lambda = 1 + \dfrac{20}{3} = \dfrac{23}{3} \)

Line equation:

\( -\dfrac{5}{3}x – 5y + \dfrac{23}{3} = 0 \)

Multiply by 3:

\( -5x – 15y + 23 = 0 \)

Pair of Straight Lines

In coordinate geometry, a second-degree equation can represent:

  • a pair of straight lines
  • a point
  • a parabola, ellipse, hyperbola

The special case where the quadratic equation represents two straight lines is very important for JEE.

1. General Second-Degree Equation

\( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \)

This represents a pair of straight lines iff:

\( abc + 2fgh – af^2 – bg^2 – ch^2 = 0 \)

This determinant-based condition is extremely important.

2. Combined Equation of Pair of Lines Through Origin

If the pair of lines passes through the origin:

\( ax^2 + 2hxy + by^2 = 0 \)

This splits into two lines:

\( (lx + my)(l’x + m’y) = 0 \)

3. Splitting the Combined Equation

If

\( ax^2 + 2hxy + by^2 = 0 \)

Divide both sides by \( x^2 \):

\( a + 2h\dfrac{y}{x} + b\dfrac{y^2}{x^2} = 0 \)

Let \( m = \dfrac{y}{x} \). Then

\( bm^2 + 2hm + a = 0 \)

Solutions \( m_1, m_2 \) give slopes of individual lines:

\( y = m_1x \), \( y = m_2x \)

4. Condition for Lines to be Real, Coincident, or Imaginary

For equation \( ax^2 + 2hxy + by^2 = 0 \):

  • Real & distinct lines if \( h^2 > ab \)
  • Coincident lines if \( h^2 = ab \)
  • Imaginary lines if \( h^2 < ab \)

5. Pair of Lines Not Through Origin

If the equation includes linear terms:

\( ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \)

Then it will represent a pair of straight lines (not necessarily through origin) only if:

\( abc + 2fgh – af^2 – bg^2 – ch^2 = 0 \)

6. Angle Between the Pair of Lines

If combined equation is

\( ax^2 + 2hxy + by^2 = 0 \)

Angle between the two lines \( \theta \) is given by:

\( \tan\theta = \dfrac{2\sqrt{h^2 – ab}}{a + b} \)

(Valid when lines are real: \( h^2 > ab \))

7. Combined Equation of Two Known Lines

If two lines are

\( L_1: a_1x + b_1y + c_1 = 0 \)
\( L_2: a_2x + b_2y + c_2 = 0 \)

Their combined equation is:

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

Example 

Show that the equation \( x^2 – 5xy + 6y^2 = 0 \) represents a pair of straight lines.

▶️ Answer / Explanation

Here:

\( a = 1,\ h = -\dfrac{5}{2},\ b = 6 \)

Check condition for pair of lines: \( h^2 \ge ab \)

\( \left(-\dfrac{5}{2}\right)^2 = \dfrac{25}{4} \)

\( ab = 6 \)

\( \dfrac{25}{4} > 6 \) → real pair of lines.

Thus it represents a pair of straight lines.

Example 

Find the slopes of the lines represented by \( 6x^2 – xy – y^2 = 0 \).

▶️ Answer / Explanation

Let \( m = \dfrac{y}{x} \).

\( -m^2 – m + 6 = 0 \)

Multiply by -1:

\( m^2 + m – 6 = 0 \)

Solve:

\( m = \dfrac{-1 \pm \sqrt{1 + 24}}{2} = \dfrac{-1 \pm 5}{2} \)

\( m_1 = 2,\quad m_2 = -3 \)

Slopes are 2 and -3.

Example 

Find the angle between the lines represented by \( 3x^2 – 7xy – 6y^2 = 0 \).

▶️ Answer / Explanation

Given:

\( a = 3,\ h = -\dfrac{7}{2},\ b = -6 \)

Formula:

\( \tan\theta = \dfrac{2\sqrt{h^2 – ab}}{a + b} \)

Compute:

\( h^2 = \dfrac{49}{4} \)

\( ab = -18 \)

\( h^2 – ab = \dfrac{49}{4} + 18 = \dfrac{49 + 72}{4} = \dfrac{121}{4} \)

\( \sqrt{h^2 – ab} = \dfrac{11}{2} \)

\( a + b = 3 – 6 = -3 \)

\( \tan\theta = \dfrac{2 \cdot \dfrac{11}{2}}{-3} = -\dfrac{11}{3} \)

Angle between lines is acute angle = \( \theta = \tan^{-1}\left(\dfrac{11}{3}\right) \)

Answer: \( \theta = \tan^{-1}\left(\dfrac{11}{3}\right) \)

Homogeneous Second Degree Equation

A homogeneous second-degree equation in two variables \( x \) and \( y \) is an equation where all terms are of total degree 2 and no linear or constant term is present.

General form:

\( ax^2 + 2hxy + by^2 = 0 \)

This is the most important special case of a conic, and in JEE this equation always represents a pair of straight lines through the origin.

1. Why Does It Represent a Pair of Lines?

Because the equation is homogeneous and degree 2, substituting \( y = mx \) reduces it to a quadratic in \( m \). Thus it splits into two lines:

\( y = m_1x \) and \( y = m_2x \)

These lines always pass through the origin \((0,0)\).

2. Condition for Nature of Lines

For equation: \( ax^2 + 2hxy + by^2 = 0 \)

  • Real and distinct lines if \( h^2 > ab \)
  • Coincident lines if \( h^2 = ab \)
  • Imaginary lines if \( h^2 < ab \)

3. Splitting into Two Lines Using Slopes

Divide equation by \( x^2 \):

\( a + 2hm + bm^2 = 0 \)

or

\( bm^2 + 2hm + a = 0 \)

Slope of lines = roots:

\( m_1, m_2 = \dfrac{-2h \pm \sqrt{4h^2 – 4ab}}{2b} = \dfrac{-h \pm \sqrt{h^2 – ab}}{b} \)

Therefore lines are:

\( y = m_1 x,\quad y = m_2 x \)

4. Angle Between the Pair of Lines

For equation:

\( ax^2 + 2hxy + by^2 = 0 \)

The angle \( \theta \) between the two lines is:

\( \tan\theta = \dfrac{2\sqrt{h^2 – ab}}{a + b} \)

This is valid only when \( h^2 > ab \) (i.e., real lines).

5. Condition for Perpendicular Lines

The pair represents perpendicular lines if:

\( a + b = 0 \)

Proof comes from the formula of \( \tan\theta \).

6. Combined Form of Two Known Lines Through Origin

If two lines passing through origin are:

\( y = m_1x \) and \( y = m_2x \)

The combined homogeneous equation is:

\( (y – m_1x)(y – m_2x) = 0 \)

Expand to get the homogeneous quadratic form.

Example

Show that \( x^2 – 5xy + 6y^2 = 0 \) represents a pair of straight lines.

▶️ Answer / Explanation

For general form: \( a = 1,\ h = -\dfrac{5}{2},\ b = 6 \)

Check condition \( h^2 > ab \).

\( h^2 = \dfrac{25}{4},\quad ab = 6 \)

\( \dfrac{25}{4} > 6 \) → real pair of lines.

Thus it represents a pair of lines through origin.

Example 

Find the two lines represented by \( 6x^2 + 5xy – 6y^2 = 0 \).

▶️ Answer / Explanation

Let \( m = \dfrac{y}{x} \).

\( 6 + 5m – 6m^2 = 0 \)

Rewrite:

\( -6m^2 + 5m + 6 = 0 \)

Multiply by -1:

\( 6m^2 – 5m – 6 = 0 \)

Use quadratic formula:

\( m = \dfrac{5 \pm \sqrt{25 + 144}}{12} = \dfrac{5 \pm 13}{12} \)

\( m_1 = \dfrac{18}{12} = \dfrac{3}{2},\quad m_2 = \dfrac{-8}{12} = -\dfrac{2}{3} \)

Thus lines are:

\( y = \dfrac{3}{2}x,\quad y = -\dfrac{2}{3}x \)

Example 

Find the angle between the lines represented by \( 4x^2 + 7xy – 2y^2 = 0 \).

▶️ Answer / Explanation

Here: \( a = 4,\ h = \dfrac{7}{2},\ b = -2 \)

Formula:

\( \tan\theta = \dfrac{2\sqrt{h^2 – ab}}{a + b} \)

\( h^2 = \dfrac{49}{4},\quad ab = -8 \)

\( h^2 – ab = \dfrac{49}{4} + 8 = \dfrac{49 + 32}{4} = \dfrac{81}{4} \)

\( \sqrt{h^2 – ab} = \dfrac{9}{2} \)

\( a + b = 4 – 2 = 2 \)

\( \tan\theta = \dfrac{2 \cdot \dfrac{9}{2}}{2} = \dfrac{9}{2} \)

\( \theta = \tan^{-1}\left(\dfrac{9}{2}\right) \)

Answer: \( \theta = \tan^{-1}\left(\dfrac{9}{2}\right) \)

Notes and Study Materials

Examples and Exercise

IIT JEE (Main) Mathematics ,”Pair of Straight Line” Notes ,Test Papers, Sample Papers, Past Years Papers , NCERT , S. L. Loney and Hall & Knight Solutions and Help from Ex- IITian

About this unit

Equations of internal and external bisectors of angles between two lines, coordinates of centroid, orthocentr, and circumcentre of a triangle, equation of family of lines passing through the point of intersection of two lines.

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