Mathematics — Std 12
📐

Pair of Straight Lines

Ch. 4Std 12

Easy Overview

In earlier years, you studied how to write the equation of a single straight line — y = mx + c, ax + by + c = 0, and so on. But what happens when two lines overlap? Or pass through the same point? Can we represent both of them with a single equation? Yes — and that's exactly what this chapter is about. A pair of straight lines is essentially two distinct lines that pass through a common point, combined into a single second-degree equation. The cool thing is that you can look at a quadratic equation in x and y and recognise whether it represents two straight lines, and if so, find their individual equations. This chapter starts with how to form the combined equation when you know the two individual lines. If L₁ = 0 and L₂ = 0 are two lines, then their product L₁ × L₂ = 0 represents both lines together. When we multiply these, we get a second-degree homogeneous equation of the form ax² + 2hxy + by² = 0 — provided both lines pass through the origin. This is the homogeneous form, and it's the simplest case. The key parameters are a, h, and b, and from these three numbers we can determine everything: whether the lines are real or imaginary, the angle between them, whether they're parallel or perpendicular, and the individual line equations. The angle between the two lines is given by the formula tan θ = |2√(h² − ab)| / |a + b|. If h² = ab, the two lines are parallel (or coincident). If a + b = 0, the lines are perpendicular. If h² < ab, no real lines exist — the pair is imaginary. So the discriminant h² − ab tells you the nature of the pair. Now, what about lines that DON'T pass through the origin? That's the general second-degree equation: ax² + 2hxy + by² + 2gx + 2fy + c = 0. This represents a pair of straight lines if and only if the determinant condition holds: Δ = abc + 2fgh − af² − bg² − ch² = 0. This is the 'pair of lines condition.' When it holds, you can shift the origin to the point of intersection of the two lines to get back to the homogeneous form. Finding the point of intersection itself involves solving ∂/∂x = 0 and ∂/∂y = 0 of the general equation — which gives the coordinates of the centre. You'll also learn how to find the separate equations of the two lines from the combined equation. The method: treat ax² + 2hxy + by² = 0 as a quadratic in x (or y), solve for x in terms of y using the quadratic formula, and factor. Each factor gives one line. Alternatively, if the general equation is given and the pair condition holds, you can factor the quadratic part, then adjust the linear terms to match the full expression. There's also a method using the formula for the slopes: the slopes m₁ and m₂ satisfy m₁ + m₂ = −2h/b and m₁m₂ = a/b (when the equation is treated as quadratic in y/x). These relationships come from the coefficients and are useful shortcuts. By the end of this chapter, you'll be able to look at any second-degree equation, quickly check if it represents a pair of lines, find the angle between them, determine their individual equations, and understand their geometric relationship. This is a short but powerful chapter — the concepts are limited but the applications range from coordinate geometry to analytic geometry used in computer vision and graphics.

Combined Equation of Two Lines

If two lines L₁: a₁x + b₁y + c₁ = 0 and L₂: a₂x + b₂y + c₂ = 0 exist, their combined equation is L₁·L₂ = 0. Multiply the two equations: (a₁x + b₁y + c₁)(a₂x + b₂y + c₂) = 0. The result is always a second-degree equation in x and y. Any point lying on either line satisfies this product equation. This is how we represent two lines with one equation.

Homogeneous Equation of Degree 2

A homogeneous equation has every term of the same degree. For degree 2: ax² + 2hxy + by² = 0. This always represents two straight lines passing through the origin (0,0). Why? Because substituting (0,0) makes the equation 0 regardless of a, h, b. The lines might be real and distinct, real and coincident, or imaginary depending on the discriminant h² − ab.

Nature of the Pair — Real or Imaginary

The discriminant determines the nature: if h² > ab, the pair consists of two distinct real lines through the origin. If h² = ab, the two lines are real and coincident (the same line repeated). If h² < ab, the pair represents no real lines — they're imaginary. This is analogous to the discriminant of a quadratic equation determining real vs. complex roots.

Angle Between Two Lines

For the pair ax² + 2hxy + by² = 0, the angle θ between the two lines is given by tan θ = |2√(h² − ab)| / |a + b|. Important special cases: if a + b = 0, then the lines are perpendicular (θ = 90°). If h² = ab, then θ = 0 — the lines are parallel or coincident. The formula assumes both lines are real (h² ≥ ab). Note the absolute values — the angle is always acute or right.

Condition for Perpendicular Lines

For the pair ax² + 2hxy + by² = 0, the lines are perpendicular if and only if a + b = 0. This is because tan 90° = ∞, which occurs when the denominator a + b = 0. This is a super useful quick test — scan the coefficients a and b. If they sum to zero, the two lines are at right angles to each other. Example: 3x² + 5xy − 3y² = 0 → a + b = 3 + (−3) = 0 → perpendicular lines.

Condition for Parallel Lines

The lines are parallel (or coincident) when h² = ab. In this case, tan θ = 0, so θ = 0°. Geometrically, both lines have the same slope. The combined equation becomes (√a x + √b y)² = 0 when h = √(ab) with the same sign, or (√a x − √b y)² = 0 when h = −√(ab). Both reduce to a single line repeated — the pair is coincident.

Slopes of the Lines from Coefficients

For ax² + 2hxy + by² = 0 (b ≠ 0), divide by x²: a(y/x)² + 2h(y/x) + b = 0. Let m = y/x be the slope. Then bm² + 2hm + a = 0. The roots m₁ and m₂ are the slopes. Sum of slopes: m₁ + m₂ = −2h/b. Product of slopes: m₁m₂ = a/b. If b = 0, the equation becomes x(ax + 2hy) = 0, giving x = 0 (vertical line) and ax + 2hy = 0. These relations are quick ways to find information without solving fully.

Finding Separate Lines from Combined Equation

Method 1: Treat ax² + 2hxy + by² = 0 as a quadratic in x. Solve for x using the quadratic formula: x = [−2hy ± √(4h²y² − 4aby²)] / (2a) = [−hy ± y√(h² − ab)] / a. This gives x = m₁y and x = m₂y, i.e., two lines through the origin. Method 2: Factor directly if the expression factors nicely. For example, x² + 5xy + 6y² = 0 → (x + 2y)(x + 3y) = 0.

General Second-Degree Equation

The general form is ax² + 2hxy + by² + 2gx + 2fy + c = 0. This represents a pair of straight lines if and only if the determinant condition holds: Δ = abc + 2fgh − af² − bg² − ch² = 0. When this condition is satisfied, the equation represents two lines that intersect at a point (not necessarily the origin). If Δ ≠ 0, it represents a conic (circle, parabola, ellipse, hyperbola) instead.

Point of Intersection of the Pair

When the general equation represents a pair of lines, they intersect at a point P(x₀, y₀). This point is found by solving ∂F/∂x = 0 and ∂F/∂y = 0, where F(x,y) is the general equation. This gives: ax₀ + hy₀ + g = 0 and hx₀ + by₀ + f = 0. Solve for x₀ and y₀. Then shifting the origin to this point converts the equation to the homogeneous form through the transformation x = X + x₀, y = Y + y₀.

Shifting Origin to Remove Linear Terms

Once the point of intersection (x₀, y₀) is found, substitute x = X + x₀, y = Y + y₀ into the general equation. The linear terms (in X and Y) vanish by construction, leaving a homogeneous equation aX² + 2hXY + bY² = 0. This is the pair of lines through the new origin. This transformation simplifies the problem to the homogeneous case we already understand.

Distance Between Two Parallel Lines

If the pair of lines are parallel (h² = ab), the general equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 (with Δ = 0) represents two parallel lines. Their distance can be found by rewriting the equation as two separate parallel line equations and using the distance formula d = |c₁ − c₂| / √(a² + b²). The lines are of the form L₁: lx + my + n₁ = 0 and L₂: lx + my + n₂ = 0.

Equation of Lines Joining Origin to Intersection Points

Given a curve (like a circle or conic) and a line, we can find the equation of the pair of lines joining the origin to the intersection points of the curve and the line. The technique: homogenise the curve equation with the help of the line equation. For example, if a circle x² + y² = r² meets the line lx + my = n, substitute using the line to homogenise the circle equation. The result is a pair of lines through the origin.

Pair of Lines Through Given Points

A pair of lines passing through two given points can be constructed. The combined equation of lines joining the origin to the points of intersection of a given line and a given curve is found by the homogenisation technique. This method is useful for problems where you need to show that certain lines through the origin form a pair with given properties.

Key Points

  • Combined equation of two lines L₁ = 0 and L₂ = 0 is L₁L₂ = 0
  • Homogeneous form ax² + 2hxy + by² = 0 always passes through origin
  • h² > ab → two distinct real lines; h² = ab → coincident lines; h² < ab → imaginary lines
  • Angle between lines: tan θ = |2√(h² − ab)| / |a + b|
  • Lines are perpendicular if and only if a + b = 0
  • Lines are parallel (or coincident) if and only if h² = ab
  • Slopes m₁, m₂ satisfy m₁ + m₂ = −2h/b and m₁m₂ = a/b (for b ≠ 0)
  • To separate lines: solve the quadratic in x or y, or factor directly
  • General second-degree equation represents a pair of lines iff Δ = 0
  • Δ = abc + 2fgh − af² − bg² − ch²
  • Point of intersection of the pair: solve ax + hy + g = 0 and hx + by + f = 0
  • Shift origin to intersection point to reduce to homogeneous form
  • Distance between parallel lines: d = |n₁ − n₂| / √(l² + m²)
  • Homogenisation: turn a curve equation into homogeneous form using a line equation
  • For perpendicular lines, a + b = 0 regardless of h
  • If b = 0 in homogeneous form, one line is x = 0 (the y-axis)

Practice Questions

  • Show that the equation x² − 5xy + 6y² = 0 represents two distinct lines. Find the angle between them.
  • Find the combined equation of the lines x + 2y − 3 = 0 and 2x − y + 1 = 0.
  • If the lines represented by ax² + 2hxy + by² = 0 are perpendicular, prove that a + b = 0.
  • Find the separate equations of the lines represented by 2x² + 5xy + 3y² = 0.
  • Show that the equation 2x² + 3xy − 2y² + 3x + y + 1 = 0 represents a pair of lines. Find their point of intersection.
  • Find the value of k such that the equation x² + kxy − 3y² + 2x + y + 1 = 0 represents a pair of lines.
  • Find the angle between the lines represented by x² + 2xy sec θ + y² = 0.
  • The slopes of the lines represented by x² + kxy + y² = 0 are in the ratio 2:3. Find k.