Notes on Straight Lines: General and POSL (JEE 2.0 2026)

Notes: Straight Lines (JEE 2.0 2026) - Tarun Khandelwal

General second-degree equation representing a pair of straight lines

• General form: $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$
• Represents a pair of straight lines iff the determinant of the coefficient matrix vanishes:
  egin{vmatrix} a & h & g \ h & b & f \ g & f & c \end{vmatrix} = 0
  which expands to $abc + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$
• If it factors, it can be written as
  $(l<em>{1}x + m</em>{1}y + n<em>{1})(l</em>{2}x + m<em>{2}y + n</em>{2}) = 0$
• Note: The equation represents two lines which may be real, coincident, parallel, or imaginary depending on coefficients.

Homogeneous second-degree equation (POSL through origin)

• Homogeneous form: $a x^{2} + 2 h x y + b y^{2} = 0$
• Represents a pair of straight lines passing through the origin (0,0).
• Slopes of the two lines given by setting $y = m x$:
  $b m^{2} + 2 h m + a = 0$
• Sums and products of slopes:
  m{1} + m{2} = -rac{2 h}{b}, \ m{1} m{2} = rac{a}{b}.
• Angle between the two lines:
  an heta = igg| rac{2 \,
  oot 4ernoulli{?}
  }{a + b} igg| = igg| rac{2 \, ext{sqrt}(h^{2} - a b)}{a + b} igg|.
  Practical form used: an heta = rac{2 \, ext{sgn}(h^{2}-ab)\,
  oot 4ernoulli{?}}{a + b} (standard result: see below).
• Important special cases:
  • Real and distinct lines if h^{2} > a b.
  • Real and coincident or parallel if $h^{2} = a b$.
  • Imaginary lines if h^{2} < a b.
  • If $a + b = 0$ then the lines are perpendicular (since $ an heta o ext{undefined}$, i.e., $ heta = 90^ ext{o}$).
• Coincident case: the homogeneous equation factors as a perfect square, $a x^{2} + 2 h x y + b y^{2} = (p x + q y)^{2} = p^{2}x^{2} + 2 p q x y + q^{2} y^{2}$, which implies $a : h : b = p^{2} : 2 p q : q^{2}.$

Angle between a pair of lines (POSL or non-POSL)

• For pair of lines represented by $a x^{2} + 2 h x y + b y^{2} = 0$ (or the general non-homogeneous case), the angle $ heta$ between them satisfies:
  $an heta = \left| \frac{2 \sqrt{h^{2} - a b}}{a + b} \right|.$
• When $a + b = 0$, $ heta = 90^ ext{o}$ (perpendicular).
• If $h^{2} > ab$, the lines are real and distinct; if $h^{2} = ab$, they may be real and coincident or parallel; if $h^{2} < ab$, they are imaginary.
• Example problem (IIT JEE 2004): The pair of lines represented by $x^{2} - y^{2} + 2y - 1 = 0$ and $x - y = 3$; the angle bisectors are found via standard bisector methods (see below).

Real/Imaginary/Parallel/Perpendicular criteria (condensed)

• For the pair $ax^{2} + 2hxy + by^{2} = 0$:
  • Distinct real lines if $h^{2} > ab$.
  • Real and coincident or parallel if $h^{2} = ab$.
  • Imaginary if $h^{2} < ab$.
• Perpendicular if $a + b = 0$ (equivalently $m{1} m{2} = -1 \, (\text{since } m{1} m{2} = a/b)$).
• Parallel iff the slopes are equal, i.e. $m{1} = m{2}$, which occurs when $h^{2} = ab$.

Distance between a pair of parallel lines

• If the pair of lines is parallel (i.e., $h^{2} = ab$) and the two lines are given by
  $l x + m y + n<em>{1} = 0, ewline l x + m y + n</em>{2} = 0,$
  then the distance between them is
  $D = \frac{|n<em>{2} - n</em>{1}|}{\sqrt{l^{2} + m^{2}}}.$

• In terms of the general second-degree coefficients, with the factorization $ax^{2} + 2hxy + by^{2} + 2g x + 2f y + c = 0$ and the relations
  g = \frac{(n{1} + n{2})}{2} l, = \frac{(n{1} + n{2})}{2} m,
  c = n{1} n{2},

  one can derive that
  D^{2} = \frac{(n{1} - n{2})^{2}}{l^{2} + m^{2}} = \frac{S^{2} - 4 c}{a + b},
  with $S = n{1} + n{2} = \frac{2 g}{l} = \frac{2 f}{m}$.
  Consequently, a practical form (using $l^{2} = a$, $m^{2} = b$, $l m = h$) is
  D = \frac{\sqrt{\frac{4 g^{2}}{a} - 4 c}}{\sqrt{a + b}} = \frac{2}{\sqrt{a + b}} \sqrt{\frac{g^{2}}{a} - c}.
  (Equivalently, $D = \frac{2}{\sqrt{a + b}} \sqrt{\frac{f^{2}}{b} - c}$, when using $f$ and $b$ form.)

Angle Bisectors for a pair of straight lines through the origin (POSL)

• For the pair of lines represented by the homogeneous equation a x^{2} + 2 h x y + b y^{2} = 0,$the combined (union) equation of the two angle bisectors is given by:<br />$\frac{x^{2} - y^{2}}{a - b} = \frac{xy}{h},\quad a + b \neq 0,\ h \neq 0.
• This is derived from the standard angle-bisector condition using normalized line equations and eliminating the common scale.
• In general, angle-bisector equations for two non-homogeneous lines L1: $l{1}x + m{1}y + n{1} = 0$ and L2: $l{2}x + m{2}y + n{2} = 0$ are given by
  \frac{l{1}x + m{1}y + n{1}}{\sqrt{l{1}^{2} + m{1}^{2}}} = \pm \frac{l{2}x + m{2}y + n{2}}{\sqrt{l{2}^{2} + m{2}^{2}}}.
• Examples in exercises include bisectors of lines represented by $4x^{2} - 16xy - 7y^{2} = 0$ (the given options show how to form the bisector equations).

Angle bisector equations for lines in general (non-homogeneous)

• If lines are $l x + m y + n = 0$ and $l' x + m' y + n' = 0$, the angle bisectors are:
  \frac{l x + m y + n}{\sqrt{l^{2} + m^{2}}} = \pm \frac{l' x + m' y + n'}{\sqrt{l'^{2} + m'^{2}}}.
• The product of the two lines is $(l x + m y + n)(l' x + m' y + n') = 0$.
• Special case used in JEE: For pair $ax^{2} + 2 h x y + by^{2} = 0$ (homogeneous), the bisectors reduce to the form above.

Locus, Concurrency, and related concepts (from notes)

• Locus, parametric forms, and the idea of POSLs (pair of lines) as locus intersections are covered in various slides.
• Parametric forms and shifting/rotation of axes are common techniques to simplify equations of straight lines and angular relationships.
• Rotation and shifting of axes help in eliminating cross-term $xy$ and simplifying the pair of lines into standard forms.

Homogenization: joining origin to intersection points (POSL)

• Idea: Given a conic (second-degree curve) ax^{2} + 2 h x y + by^{2} + 2 g x + 2 f y + c = 0 and a line $L: lx + my + n = 0$, the homogenization of the curve with respect to L yields a homogeneous equation whose zero-set represents the two lines joining the origin to the points where the curve meets L.
• The standard homogenization in projective coordinates uses a third variable $z$ and writes the conic as
  a x^{2} + 2 h x y + b y^{2} + 2 g x z + 2 f y z + c z^{2} = 0,
  and the reference line becomes $L(x,y,z)=0$ with $z=0$ representing the line at infinity in the homogenized space.
• When $z=0$, the homogeneous part of degree 2 reduces to the pair of lines ax² + 2hxy + by² = 0, which pass through the origin, corresponding to lines joining the origin to the intersection points of the original curve with the line L.
• Worked example (from notes): Find the equation of the pair of lines joining the origin to the intersection points of the line $2x - y = 3$ and the curve $x^{2} - y^{2} - xy + 3x - 6y + 18 = 0$.
  • The homogenized form with respect to $L: 2x - y - 3 z = 0$ yields a degree-2 homogeneous equation whose factorization gives the two lines through the origin. The resulting pair is:
    11 x^{2} + 3 y^{2} - 14 x y = 0.
  • The angle between these two lines can then be computed using the slope-based method or the angle formula above.

Problems and applications (selected concepts from notes)

• IIT JEE 2004 style problem: Area of the triangle formed by the angle bisectors of the pair of lines and a given line can be solved using angle-bisector equations and standard geometry relations.
• Examples also include determining whether two lines are perpendicular, parallel, or coincident from a general second-degree equation, using the discriminant $h^{2} - ab$ and the sum $a + b$.
• Problems often require using homogenization to handle lines through the origin or to produce POSLs that pass through the origin.
• A common technique is to substitute $y = m x$ to find slopes and then apply relationships between sums and products of slopes to extract coefficients or to solve for unknown parameters like $ ext{tan} heta$ or $ an^{-1}$ of the angle.

Quick reference: essential formulas (summary)

• Pair of lines condition: egin{vmatrix} a & h & g \ h & b & f \ g & f & c \end{vmatrix} = 0 \ ext{(equivalently } abc + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0 ext{)}$</li> <li>Homogeneous pair:$a x^{2} + 2 h x y + b y^{2} = 0 \ (m{1} + m{2}) = -\frac{2h}{b}, \ m{1}m{2} = \frac{a}{b}$</li> <li>Angle between POSL:$\tan \theta = \left| \frac{2 \sqrt{h^{2} - a b}}{a + b} \right|$</li> <li>Parallel condition:$h^{2} = ab \ \ ext{Perpendicular condition: } a + b = 0 \ \ ext{Coincident if } a = p^{2}, h = p q, b = q^{2} \text{ for some } p,q
• Distance between parallel lines (in form $l x + m y + n = 0$): D = \frac{|n{2} - n{1}|}{\sqrt{l^{2} + m^{2}}}$</li> <li>Distance in terms of coefficients (when usable):$D = \frac{2}{\sqrt{a + b}} \sqrt{\frac{g^{2}}{a} - c} (or equivalently with $f$ and $b$ under appropriate normalization)
• Angle bisectors (POSL): \frac{x^{2} - y^{2}}{a - b} = \frac{xy}{h}, \quad a + b \neq 0,\ h \neq 0$</li> <li>Combined angle bisectors for non-homogeneous lines:$\frac{l x + m y + n}{\sqrt{l^{2} + m^{2}}} = \pm \frac{l' x + m' y + n'}{\sqrt{l'^{2} + m'^{2}}}.
• Homogenization principle: given a curve $ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0$ and a line $lx + my + n = 0$, the homogenized form in coordinates $(x,y,z)$ is
  a x^{2} + 2h x y + b y^{2} + 2g x z + 2f y z + c z^{2} = 0,$$
  and the POSL (origin-connected lines) arise from the intersection when $z=0$.
• A notable result: The angle between lines joining the origin to the intersection points of a line and a conic can be studied via homogenization and factoring into a pair of lines through the origin.

Note: Several worked examples in the notes revolve around applying the above formulas to specific line-equation pairs, angle-bisector problems, and homogenization exercises to obtain the POSL (pair of lines through the origin) and to compute angles and distances between such lines.