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State Board · Class 12 · Mathematics Part 1

Lesson 4: Pair of Straight Lines

3 concepts · 24 practice questions with answers · Free to read

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Homogeneous equation — two lines through the origin

From daily life

Two roads start from the same crossing, which is the origin, and one joint equation can describe both. Building on class 11 : in class 11 you met one line y = mx; here we take two lines together.

Simple definition

The homogeneous second-degree equation ax² + 2hxy + by² = 0 represents two lines through the origin (real if h² ≥ ab). Putting y = mx gives bm² + 2hm + a = 0.

Rule 1: Sum and product of slopes — m₁ + m₂ = −2h/b; m₁m₂ = a/b.

Rule 2: Forming the joint equation — For lines L₁ = 0 and L₂ = 0 the joint equation is L₁L₂ = 0.

y = m₁xy = m₂xyx

In the figure : The two lines meet at the origin.

How is the formula obtained? : x² − 5xy + 6y² = 0 ⇒ (x − 2y)(x − 3y) = 0, so the lines are x = 2y and x = 3y with slopes ½ and ⅓; the sum 5/6 = −2h/b and the product 1/6 = a/b.

  • Easy : x² − y² = 0 ⇒ (x − y)(x + y) = 0, so y = x and y = −x.
  • Medium : For y = 2x and y = −x: (y − 2x)(y + x) = y² − xy − 2x² = 0, i.e. 2x² + xy − y² = 0.
  • Exam-type : 2x² + xy − y² = 0 ⇒ a = 2, 2h = 1, b = −1 ⇒ product of slopes a/b = −2 (y = 2x, y = −x: 2 × (−1) = −2).
Remember

"Homogeneous" means every term has degree 2.

Common mistake

Taking 2h as h — the coefficient of the middle term is 2h.

Shortcut

If it factorises, factorising is faster than the formulas. Memory clue : "a, 2h, b — the middle one is doubled".

CET trap

Reduce to integer coefficients before reading a, h, b.

CET-style example : The lines of x² − 5xy + 6y² = 0 are: (A) x = 2y, x = 3y (B) x = −2y, x = −3y (C) y = 2x, y = 3x (D) x = y, x = 6y

Answer : (A). Time-saving method : Factors (x − 2y)(x − 3y).

Spot the difference

अटCondition
h² > abh² > ab
h² = abh² = ab
h² < abh² < ab

Practice questions

Easy
  1. 1Write the lines of x² − 4y² = 0.
  2. 2Joint equation of y = x and y = 3x?
  3. 3For ax² + 2hxy + by² = 0 what is m₁ + m₂?
Medium
  1. 4Find the slopes for x² − 5xy + 6y² = 0.
  2. 5Lines of 2x² + xy − y² = 0?
  3. 6Here h² = ab for x² + 2xy + y² = 0. What does it mean?
Challenging
  1. 7If the product of slopes is −1, what is the relation between a and b?
  2. 8Write the lines of x² − 5xy + 6y² = 0 in the form y = mx.
Show answers (try them yourself first)
  1. 1x = 2y and x = −2y.
  2. 2(y − x)(y − 3x) = y² − 4xy + 3x² = 0.
  3. 3−2h/b.
  4. 4½ and ⅓.
  5. 5(2x − y)(x + y) = 0 ⇒ y = 2x, y = −x.
  6. 6h = 1, ab = 1 ⇒ (x + y)² = 0, both lines coincide.
  7. 7a/b = −1 ⇒ a + b = 0.
  8. 8y = ½x and y = ⅓x.

Angle between the lines and special conditions

From daily life

Once the angle between two roads at a crossing is known, we can say whether they are perpendicular or merged.

Simple definition

The acute angle θ between the lines of ax² + 2hxy + by² = 0 is given by tan θ = |2√(h² − ab) / (a + b)|. Special cases: a + b = 0 means perpendicular; h² = ab means coincident (θ = 0).

Rule 1: Angle formula — tan θ = |2√(h² − ab)/(a + b)|.

Rule 2: Perpendicular lines — a + b = 0 (coefficients of x² and y² are opposite).

Rule 3: Coincident lines — h² = ab.

How is the formula obtained? : x² − 5xy + 6y² = 0: a = 1, h = −5/2, b = 6, so h² − ab = 25/4 − 6 = 1/4 and tan θ = 2(½)/7 = 1/7; directly from slopes: (½ − ⅓)/(1 + 1/6) = 1/7.

  • Easy : x² − y² = 0: a + b = 1 − 1 = 0, so the lines are perpendicular.
  • Medium : x² + kxy + 9y² = 0 coincident: h = k/2 and h² = ab ⇒ k²/4 = 9 ⇒ k = ±6.
  • Exam-type : kx² + 5xy − 3y² = 0 perpendicular: k + (−3) = 0 ⇒ k = 3.
Remember

Do not use the angle formula when a + b = 0 (zero denominator): the angle is 90°.

Common mistake

Using the coefficient itself as h instead of half of it.

Shortcut

See a + b = 0 and answer 90° at once. Memory clue : "Sum zero = right angle; h² = ab = one line".

CET trap

The angle is acute; take the absolute value if tan comes out negative.

CET-style example : If the lines of 3x² + 2hxy + 4y² = 0 coincide, h = ? (A) ±√12 (B) ±12 (C) ±3 (D) ±4

Answer : (A). h² = 3·4 = 12. Time-saving method : Directly h² = ab.

Spot the difference

अर्थMeaning
लंब रेषाperpendicular lines
संपातीcoincident
दोन वेगळ्याtwo distinct

Practice questions

Easy
  1. 1Angle between the lines of x² − y² = 0?
  2. 2Angle between the lines of x² + 2xy + y² = 0?
  3. 3tan θ for x² − 5xy + 6y² = 0?
Medium
  1. 4If kx² + 5xy − 3y² = 0 gives perpendicular lines, find k.
  2. 5If x² + kxy + 9y² = 0 gives coincident lines, find k.
  3. 6Find tan of the angle for 2x² + 3xy + y² = 0.
Challenging
  1. 7If 3x² + 2hxy + 4y² = 0 is coincident, find h.
  2. 8Write both lines of 2x² + 3xy + y² = 0.
Show answers (try them yourself first)
  1. 190° (a + b = 0).
  2. 20° (coincident).
  3. 31/7.
  4. 4k = 3.
  5. 5k = ±6.
  6. 6h = 3/2, h² − ab = 9/4 − 2 = 1/4, so tan θ = 2(½)/3 = 1/3.
  7. 7h² = 12 ⇒ h = ±2√3.
  8. 8(2x + y)(x + y) = 0 ⇒ y = −2x, y = −x.

General second-degree equation as a pair of lines

From daily life

Not every pair of lines passes through the origin; two fence wires can lie anywhere. Their equation is a complete second-degree one.

Simple definition

The equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines if abc + 2fgh − af² − bg² − ch² = 0 (or the determinant [[a, h, g], [h, b, f], [g, f, c]] = 0). The point of intersection is found by solving the two lines.

Rule 1: Condition — abc + 2fgh − af² − bg² − ch² = 0.

Rule 2: Angle — The angle depends only on the x², xy, y² part: tan θ = |2√(h² − ab)/(a + b)|.

How is the formula obtained? : 2x² + 7xy + 3y² + 8x + 14y + λ = 0: a = 2, h = 7/2, b = 3, g = 4, f = 7, c = λ ⇒ 6λ + 196 − 98 − 48 − λ·49/4 = 0 ⇒ 50 − 6.25λ = 0 ⇒ λ = 8. The lines are (2x + y + 4)(x + 3y + 2) = 0.

  • Easy : x² − y² + 2x + 1 = (x + 1)² − y² = 0 ⇒ x + 1 = y and x + 1 = −y; the condition gives 0.
  • Medium : Intersection of the above lines: x + 1 = 0 and y = 0 ⇒ (−1, 0).
  • Exam-type : Intersection of 2x + y + 4 = 0 and x + 3y + 2 = 0: x = −2, y = 0.
Remember

The determinant must be 0 for a pair of lines.

Common mistake

Taking g and f as 2g and 2f; the formula uses half the coefficients.

Shortcut

If you can factorise, that itself proves it is a pair of lines. Memory clue : "abc + 2fgh − af² − bg² − ch² = 0".

CET trap

Remember g and f are half the coefficients of x and y.

CET-style example : Does x² − y² + 2x + 1 = 0 represent a pair of lines? (A) Yes (B) No (C) Only for x (D) Cannot say

Answer : (A). (x + 1)² − y². Time-saving method : Try factorising first.

Spot the difference

ओळखTest
h² ≥ ab पुरेh² ≥ ab is enough
abc + 2fgh − af² − bg² − ch² = 0abc + 2fgh − af² − bg² − ch² = 0

Practice questions

Easy
  1. 1Is x² − y² + 2x + 1 = 0 a pair of lines?
  2. 2Point of intersection of those lines?
  3. 3What does x² + y² + 2x + 2y + 2 = 0 represent?
Medium
  1. 4Expand (2x + y + 4)(x + 3y + 2).
  2. 5If 2x² + 7xy + 3y² + 8x + 14y + λ = 0 is a pair of lines, find λ.
  3. 6Their point of intersection?
Challenging
  1. 7Product of the slopes of those lines?
  2. 8tan of the acute angle between those lines?
Show answers (try them yourself first)
  1. 1Yes: (x + 1 − y)(x + 1 + y) = 0.
  2. 2(−1, 0).
  3. 3(x + 1)² + (y + 1)² = 0, so only the point (−1, −1).
  4. 42x² + 7xy + 3y² + 8x + 14y + 8.
  5. 5λ = 8.
  6. 6(−2, 0).
  7. 7Slopes −2 and −1/3, product 2/3 = a/b.
  8. 8|(−2 + 1/3)/(1 + 2/3)| = |(−5/3)/(5/3)| = 1, so 45°.