Pair of Straight Lines — MHT-CET Maths Formulas and Traps

Joint Equation of Two Lines — Product of Linear Factors and the Triangle They Form

The Joint Equation Is the Product: (L₁)(L₂) = 0

Joint equation

(a_1x + b_1y + c_1)(a_2x + b_2y + c_2) = 0 \qquad \text{through the origin: } (y - m_1x)(y - m_2x) = 0

Factorising a Pair: Split the Middle Term, or Complete the Square

Factorising

ax^2 + 2hxy + by^2 = b(y - m_1x)(y - m_2x) \qquad \text{grouping: } xy + px + qy + pq = (x + q)(y + p)

The Triangle a Pair Makes With a Third Line: Vertices, Centroid, Median, Circumcentre

Triangle from a pair

\text{vertices: } O,\ L_1 \cap L_3,\ L_2 \cap L_3;\qquad \text{centroid} = \frac{\sum \text{vertices}}{3}

Common traps

30° to the Y-axis read as slope tan 30°

A line at

30^\circ

to the

Y

-axis is at

60^\circ

to the

X

-axis: slope

\sqrt3

, joint equation

3x^2 - y^2 = 0

. Option

x^2 - 3y^2 = 0

is the

\tan 30^\circ

slip.

Reading xy − x + y − 1 as a curve

Any equation of the form

xy + px + qy + pq = 0

is two lines. Grouping finds them; treating it as a hyperbola sends you down the wrong chapter.

Finding A and B explicitly for the median

The median from

O

needs only the MIDPOINT of

AB

, which Vieta gives from the quadratic without solving it. Solving

37y^2 - 14y + 1 = 0

by formula wastes the question's time budget.

Slopes of a Homogeneous Pair — Sum, Product and Ratio Conditions

Learn this subtopic in the notes

m₁ + m₂ = −2h/b and m₁m₂ = a/b

Vieta for slopes

bm^2 + 2hm + a = 0:\qquad m_1 + m_2 = -\frac{2h}{b},\qquad m_1 m_2 = \frac{a}{b}

Slopes in a Ratio: (m + n)² ab = 4mn h², and the Reverse Direction

Ratio identity

m_1 : m_2 = m : n \iff (m + n)^2\,ab = 4mn\,h^2 \qquad \frac{(m_1 + m_2)^2}{m_1 m_2} = \frac{4h^2}{ab}

A Common Line Between Two Pairs, and a Line of the Pair Perpendicular to a Given Line

Membership test

y = kx \text{ belongs to } ax^2 + 2hxy + by^2 = 0 \iff a + 2hk + bk^2 = 0

Common traps

Dividing by a instead of b

m_1 m_2 = \dfrac{a}{b}

, the

x^2

coefficient over the

y^2

coefficient. Inverting it gives

K = \dfrac{4}{27}

, option (D) on the

Kx^2 + 6xy + y^2

stem.

Writing the identity with h² and ab swapped

(m + n)^2 ab = 4mn h^2

, so

ab : h^2 = 8 : 9

for the ratio

1 : 2

, not

9 : 8

. Both orders are always offered; check with

m = n = 1

, which must give

h^2 = ab

.

Substituting the given line's own slope

Perpendicular to

mx + ny = 18

means slope

\dfrac{n}{m}

, which gives

am^2 + 2hmn + bn^2 = 0

. Using

-\dfrac{m}{n}

swaps

m

and

n

and flips a sign — options (A) and (D) on that stem.

Angle Between the Pair — Perpendicular Pairs, Lines at a Given Angle and the Bisectors

Learn this subtopic in the notes

tan θ = 2√(h² − ab)/|a + b|: Perpendicular When a + b = 0, Parallel When h² = ab

Angle between the pair

\tan\theta = \frac{2\sqrt{h^2 - ab}}{|a + b|} \qquad a + b = 0 \iff \perp \qquad h^2 = ab \iff \parallel

The Pair Through a Point at a Given Angle to a Line: Square the Angle Condition

Pair at angle α to ax + by = 0

(ax + by)^2 = \tan^2\alpha\,(bx - ay)^2 \qquad \text{perpendicular pair to } ax^2 + 2hxy + by^2: \ bx^2 - 2hxy + ay^2 = 0

The Angle Bisectors: (x² − y²)/(a − b) = xy/h

Bisector pair

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

Common traps

Using |a − b| in the denominator

The denominator is

|a + b|

;

a - b

belongs to the BISECTOR formula. Mixing them turns a

60^\circ

answer into

\tan^{-1}

of something not on the list — or worse, onto a distractor.

Sign of the xy term after substituting m = y/x

5m^2 - 24m - 5 = 0

becomes

5y^2 - 24xy - 5x^2 = 0

; multiplying by

-1

gives

5x^2 + 24xy - 5y^2 = 0

. The three wrong options differ only in these signs; keep the substitution explicit.

Using the full xy coefficient as h

For

2x^2 + 11xy + 3y^2 = 0

,

h = \dfrac{11}{2}

. Using

11

gives

11x^2 + 4xy - 11y^2

-type answers that miss every option by a factor in the middle term.

General Second-Degree Equation — Condition for a Pair, Parallel Lines and Distances

Learn this subtopic in the notes

Condition for a Pair: abc + 2fgh − af² − bg² − ch² = 0

Pair condition

\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = abc + 2fgh - af^2 - bg^2 - ch^2 = 0

Parallel Pair (h² = ab): Factor as a Perfect Square, or Use 2√((g² − ac)/(a(a + b)))

Parallel pair

h^2 = ab:\quad d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}} = 2\sqrt{\frac{f^2 - bc}{b(a + b)}}

Product of the Perpendicular Distances From a Point to the Two Lines

Product of distances

P_1 P_2 = \frac{|ax_1^2 + 2hx_1y_1 + by_1^2|}{\sqrt{(a - b)^2 + 4h^2}}

Common traps

Counting k = 0 as a pair

k = 0

satisfies

\Delta = 0

but reduces the equation to

10x + 8y + 16 = 0

, a single line. 'k = 0 or 5' is option (C) and wrong.

Reporting the gap squared, or halving it

The formula already carries the factor

2

:

2\sqrt{16/400} = \dfrac25

.

\dfrac15

(no

2

) and

5

(squared reciprocal) are the neighbours on the list.

Adding the distances

The stem asks for the PRODUCT

P_1 P_2

. The sum

\sqrt5 + \dfrac{4}{\sqrt5} = \dfrac{9}{\sqrt5}

is not offered, but the squared product

16

-style and

5

,

10

are.

Pair of Straight Lines formulas

Joint Equation of Two Lines — Product of Linear Factors and the Triangle They Form

The Joint Equation Is the Product: (L₁)(L₂) = 0

Factorising a Pair: Split the Middle Term, or Complete the Square

The Triangle a Pair Makes With a Third Line: Vertices, Centroid, Median, Circumcentre

Common traps

Slopes of a Homogeneous Pair — Sum, Product and Ratio Conditions

m₁ + m₂ = −2h/b and m₁m₂ = a/b

Slopes in a Ratio: (m + n)² ab = 4mn h², and the Reverse Direction

A Common Line Between Two Pairs, and a Line of the Pair Perpendicular to a Given Line

Common traps

Angle Between the Pair — Perpendicular Pairs, Lines at a Given Angle and the Bisectors

tan θ = 2√(h² − ab)/|a + b|: Perpendicular When a + b = 0, Parallel When h² = ab

The Pair Through a Point at a Given Angle to a Line: Square the Angle Condition

The Angle Bisectors: (x² − y²)/(a − b) = xy/h

Common traps

General Second-Degree Equation — Condition for a Pair, Parallel Lines and Distances

Condition for a Pair: abc + 2fgh − af² − bg² − ch² = 0

Parallel Pair (h² = ab): Factor as a Perfect Square, or Use 2√((g² − ac)/(a(a + b)))

Product of the Perpendicular Distances From a Point to the Two Lines

Common traps

More MHT-CET Maths formula sheets