Maharashtra State Board · Class 12 · Mathematics · Part I Chapter 4

Pair of Straight Lines — Formula Sheet

Board Formulas

14 formulas

1.Combined Equation of Two Lines
$(a_{1} x + b_{1} y + c_{1}) (a_{2} x + b_{2} y + c_{2}) = 0$
Multiply the two equations written in the form u = 0, v = 0. A point lies on the pair if it lies on at least one of the lines.
2.Pair of Lines through Origin from Slopes
$(y - m_{1} x) (y - m_{2} x) = 0 ⟺ m_{1} m_{2} x^{2} - (m_{1} + m_{2}) x y + y^{2} = 0$
$m_{1}, m_{2}$ : Slopes of the two lines
Lines y = m₁x and y = m₂x. Expanding the product gives a homogeneous equation of degree two.
3.Homogeneous Equation of Degree Two
$a x^{2} + 2 h x y + b y^{2} = 0 represents two lines through the origin ⟺ h^{2} - ab \geq 0$
$a, b$ : Coefficients of x² and y² · $h$ : Half the coefficient of xy
Every term has degree two, so (0, 0) always satisfies it. If h² − ab < 0, only the origin satisfies the equation — there are no real lines.
4.Auxiliary Equation in m
$y = m x \Rightarrow b m^{2} + 2 hm + a = 0$
Its two roots are the slopes m₁ and m₂ of the lines (when b ≠ 0). If b = 0, one line is x = 0.
5.Sum and Product of Slopes★
$m_{1} + m_{2} = - \frac{2 h}{b}, m_{1} m_{2} = \frac{a}{b} (b \neq = 0)$
Sum = −(coefficient of xy)/(coefficient of y²); product = (coefficient of x²)/(coefficient of y²).
6.Nature of the Lines
$h^{2} > ab : real and distinct, h^{2} = ab : coincident, h^{2} < ab : no real lines$
Coincident means the equation is a perfect square, e.g. x² − 6xy + 9y² = (x − 3y)².
7.Acute Angle between the Lines★
$tan θ = \frac{2 h ^{2} - ab}{a + b}$
$θ$ : Acute angle between the two lines
θ is the acute angle. If a + b = 0 the formula breaks down because θ = 90°.
8.Condition for Perpendicular Lines
$a + b = 0$
Coefficient of x² + coefficient of y² = 0, e.g. 3x² + 8xy − 3y² = 0. The value of h does not matter.
9.Condition for Coincident Lines
$h^{2} = ab$
For ax² + 2hxy + by² = 0: then tan θ = 0, so the two lines coincide. For the general second-degree equation, h² = ab means the two lines are parallel.
10.Pair through Origin Perpendicular to a Given Pair★
$a x^{2} + 2 h x y + b y^{2} = 0 ⟶ b x^{2} - 2 h x y + a y^{2} = 0$
Swap the coefficients of x² and y² and change the sign of the xy term. Check: the new slopes are −1/m₁ and −1/m₂.
11.General Equation of Second Degree
$a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0$
$g, f$ : Half the coefficients of x and y · $c$ : Constant term
Read the coefficients in this form: half of the xy, x and y coefficients give h, g and f.
12.Necessary Conditions for a Pair of Lines★
$a h g h b f g f c = 0 ⟺ ab c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0, h^{2} - ab \geq 0$
The textbook calls these necessary conditions: use them to find an unknown such as k. To show that an equation does represent a pair of lines, factorise it into two linear factors. The determinant is linear in c, so 'find k' questions with k as the constant are quick.
13.Point of Intersection of the Lines
$a x + h y + g = 0, h x + b y + f = 0 \Rightarrow (x, y) = (\frac{h f - b g}{ab - h ^{2}}, \frac{g h - a f}{ab - h ^{2}})$
Solving the two linear equations is usually easier than remembering the formula. Requires ab − h² ≠ 0 (lines not parallel).
14.Lines through Origin Parallel to the Pair
$a x^{2} + 2 h x y + b y^{2} = 0$
The second-degree part alone gives two lines through the origin parallel to the given pair, so the angle formula tan θ = |2√(h² − ab)/(a + b)| also applies to the general equation.