Pair of Straight Lines — MHT-CET Maths

Formulas (3)

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

Watch out for (3)

• 30° to the Y-axis read as slope tan 30°
  → The Joint Equation Is the Product: (L₁)(L₂) = 0
• Reading xy − x + y − 1 as a curve
  → Factorising a Pair: Split the Middle Term, or Complete the Square
• Finding A and B explicitly for the median
  → The Triangle a Pair Makes With a Third Line: Vertices, Centroid, Median, Circumcentre

Formulas (3)

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

Watch out for (3)

• Dividing by a instead of b
  → m₁ + m₂ = −2h/b and m₁m₂ = a/b
• Writing the identity with h² and ab swapped
  → Slopes in a Ratio: (m + n)² ab = 4mn h², and the Reverse Direction
• Substituting the given line's own slope
  → A Common Line Between Two Pairs, and a Line of the Pair Perpendicular to a Given Line

Formulas (3)

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

Watch out for (3)

• Using |a − b| in the denominator
  → tan θ = 2√(h² − ab)/|a + b|: Perpendicular When a + b = 0, Parallel When h² = ab
• Sign of the xy term after substituting m = y/x
  → The Pair Through a Point at a Given Angle to a Line: Square the Angle Condition
• Using the full xy coefficient as h
  → The Angle Bisectors: (x² − y²)/(a − b) = xy/h

Formulas (3)

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

Watch out for (3)

• Counting k = 0 as a pair
  → Condition for a Pair: abc + 2fgh − af² − bg² − ch² = 0
• Reporting the gap squared, or halving it
  → Parallel Pair (h² = ab): Factor as a Perfect Square, or Use 2√((g² − ac)/(a(a + b)))
• Adding the distances
  → Product of the Perpendicular Distances From a Point to the Two Lines