Is it correct to say: Taking h as the full coefficient of xy?

The xy term is 2hxy. In 2x² + 5xy + 2y² = 0, h = 5/2.

Is it correct to say: Writing the sum of slopes as 2h/b or the product as b/a?

m₁ + m₂ = −2h/b and m₁m₂ = a/b. Derive them from bm² + 2hm + a = 0 if in doubt.

Is it correct to say: Forgetting the modulus in the angle formula and reporting an obtuse angle as the acute one?

tan θ = |2√(h² − ab)/(a + b)| gives the acute angle; the obtuse angle is π − θ.

Is it correct to say: Writing the perpendicular pair as ax² − 2hxy + by² = 0?

Swap a and b as well: the pair perpendicular to ax² + 2hxy + by² = 0 is bx² − 2hxy + ay² = 0.

Is it correct to say: Checking only the determinant condition for a general second-degree equation?

Also check h² − ab ≥ 0; otherwise the equation does not give real lines.

Is it correct to say: Mixing up g and f?

g goes with x (2gx) and f goes with y (2fy). The determinant rows are (a, h, g), (h, b, f), (g, f, c).

📄 Formula sheet (PDF)

Pair of Straight Lines

Combined equation of two lines, homogeneous equation of degree two, sum and product of slopes, angle between the lines and the condition for a general second-degree equation to represent a pair of lines — Maharashtra HSC Maths Part I Ch 4

📐 14 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→For any question about slopes, put y = mx in the homogeneous equation to get bm² + 2hm + a = 0, then use the sum and product of its roots.
→Read a, h and b carefully: the coefficient of xy is 2h, so in x² − 4xy + y² = 0, h = −2 (not −4).
→For the condition on a general second-degree equation, write the 3 × 3 determinant with a, h, g / h, b, f / g, f, c — it is easier to expand correctly than the long formula.
→To find separate equations of a non-homogeneous pair, factorise the second-degree part first, then fix the constants by comparing the x, y and constant terms.
→Perpendicular lines: a + b = 0. Coincident lines: h² = ab. State the condition before substituting.

📐 Formulas(14)

Combined Equation of Two Lines

(a_{1} x + b_{1} y + c_{1}) (a_{2} x + b_{2} y + c_{2}) = 0

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

Symbol	Meaning
$m_{1}, m_{2}$	Slopes of the two lines

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

Symbol	Meaning
$a, b$	Coefficients of x² and y²
$h$	Half the coefficient of xy

Auxiliary Equation in m

y = m x \Rightarrow b m^{2} + 2 hm + a = 0

Sum and Product of Slopes★ Board fav

m_{1} + m_{2} = - \frac{2 h}{b}, m_{1} m_{2} = \frac{a}{b} (b \neq = 0)

Nature of the Lines

h^{2} > ab : real and distinct, h^{2} = ab : coincident, h^{2} < ab : no real lines

Acute Angle between the Lines★ Board fav

tan θ = \frac{2 h ^{2} - ab}{a + b}

Symbol	Meaning
$θ$	Acute angle between the two lines

Condition for Perpendicular Lines

a + b = 0

Condition for Coincident Lines

h^{2} = ab

Pair through Origin Perpendicular to a Given Pair★ Board fav

a x^{2} + 2 h x y + b y^{2} = 0 ⟶ b x^{2} - 2 h x y + a y^{2} = 0

General Equation of Second Degree

a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

Symbol	Meaning
$g, f$	Half the coefficients of x and y
$c$	Constant term

Necessary Conditions for a Pair of Lines★ Board fav

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

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}})

Lines through Origin Parallel to the Pair

a x^{2} + 2 h x y + b y^{2} = 0

✏️ Solved Examples

1Solved Exampleeasy3 steps

Find the combined equation of the lines x + 2y = 0 and 3x − y = 0.

Multiply the left-hand sides.

(x + 2 y) (3 x - y) = 0

2Solved Exampleboard4 steps

Find the acute angle between the lines represented by x² − 4xy + y² = 0.

Compare with ax² + 2hxy + by² = 0. Here 2h = −4.

a = 1, h = - 2, b = 1

3Solved Exampleboard5 steps

Find c if 3x² − 10xy − 8y² + 14x − 14y + c = 0 represents a pair of lines. Then find their point of intersection.

Compare with the general equation: 2h = −10, 2g = 14, 2f = −14.

a = 3, h = - 5, b = - 8, g = 7, f = - 7

4Solved ExampleHOTS4 steps

Find k if the slope of one of the lines given by kx² + 6xy − y² = 0 exceeds the slope of the other by 10.

Put y = mx and divide by x².

k + 6 m - m^{2} = 0 \Rightarrow m^{2} - 6 m - k = 0

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Taking h as the full coefficient of xy
✓The xy term is 2hxy. In 2x² + 5xy + 2y² = 0, h = 5/2.
2
Writing the sum of slopes as 2h/b or the product as b/a
✓m₁ + m₂ = −2h/b and m₁m₂ = a/b. Derive them from bm² + 2hm + a = 0 if in doubt.
3
Forgetting the modulus in the angle formula and reporting an obtuse angle as the acute one
✓tan θ = |2√(h² − ab)/(a + b)| gives the acute angle; the obtuse angle is π − θ.
4
Writing the perpendicular pair as ax² − 2hxy + by² = 0
✓Swap a and b as well: the pair perpendicular to ax² + 2hxy + by² = 0 is bx² − 2hxy + ay² = 0.
5
Checking only the determinant condition for a general second-degree equation
✓Also check h² − ab ≥ 0; otherwise the equation does not give real lines.
6
Mixing up g and f
✓g goes with x (2gx) and f goes with y (2fy). The determinant rows are (a, h, g), (h, b, f), (g, f, c).

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the joint equation of the pair of lines through the origin perpendicular to the lines 2x² − 3xy − 9y² = 0.
Q2
Find the separate equations of the lines represented by x² − 5xy + 6y² = 0.
Q3
Find the acute angle between the lines 2x² + 5xy + 2y² = 0.
Q4
Find k if the lines represented by 3x² + kxy + 3y² = 0 are coincident.
Q5
Show that 2x² + xy − y² + x + 4y − 3 = 0 represents a pair of lines and find their separate equations.
Q6
Find the acute angle between the lines represented by 3x² − 10xy − 8y² + 14x − 14y + 15 = 0.

📝 Notes

Pair of Straight Lines

One second-degree equation can describe two straight lines at once. The chapter moves from the simple case — two lines through the origin — to the general second-degree equation, and the same few coefficients (a, h, b, and then g, f, c) control everything.

The homogeneous case

For $ax^2 + 2hxy + by^2 = 0$ , putting $y = mx$ turns the question into a quadratic in the slope: $bm^2 + 2hm + a = 0$ . Every standard result follows from its roots:

sum of slopes $-\frac{2h}{b}$ and product $\frac{a}{b}$ ;
acute angle $\tan\theta = \left|\frac{2\sqrt{h^2 - ab}}{a + b}\right|$ ;
perpendicular when $a + b = 0$ , coincident when $h^2 = ab$ .

Questions of the form 'one slope is three times the other' or 'slopes differ by 4' are solved by writing $m_1$ and $m_2$ in terms of one unknown and using the sum and product.

The general second-degree equation

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines only when the determinant with rows $(a, h, g)$ , $(h, b, f)$ , $(g, f, c)$ is zero and $h^2 - ab \ge 0$ . These conditions are necessary, so use them to find an unknown; to show that an equation does represent a pair of lines, factorise it. The determinant is linear in a, b and c, so when k sits in one of those places the 'find k' equation is linear (k in h, g or f gives a quadratic).

The second-degree part $ax^2 + 2hxy + by^2 = 0$ gives lines through the origin parallel to the pair, so the angle formula is unchanged. The point of intersection comes from solving $ax + hy + g = 0$ and $hx + by + f = 0$ .

Finding separate equations

Factorise the second-degree part into two linear factors, write the pair as $(l_1x + m_1y + p)(l_2x + m_2y + q) = 0$ , expand, and compare the coefficients of x, y and the constant term to find p and q.