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Mathematics · Ch 17 — Pair of Straight Lines

General Second-Degree Equation Representing a Pair of Lines

17.6

General Second-Degree Equation Representing a Pair of Lines

So far every pair of lines considered has passed through the origin, giving a purely homogeneous equation. In general, a pair of straight lines anywhere in the plane is represented by the general second-degree equation

ax2+2hxy+by2+2gx+2fy+c=0,ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0,

but not every such equation actually splits into two straight lines — most represent a genuine conic (an ellipse, parabola, or hyperbola) instead. We need a condition on a,h,b,g,f,ca,h,b,g,f,c that decides when the equation does factor into two linear expressions.

Treat the equation as a quadratic in xx (assuming a≠0a\neq0):

ax2+2(hy+g)x+(by2+2fy+c)=0.ax^2 + 2(hy+g)x + (by^2+2fy+c) = 0.

For this to split into two linear factors of the form (…)(…)(\ldots)(\ldots) — each linear in both xx and yy — the value of xx obtained from the quadratic formula must be a linear expression in yy, not one involving a genuine square root of yy. That is, the discriminant

(hy+g)2−a(by2+2fy+c)(hy+g)^2 - a(by^2+2fy+c)

must itself be a perfect square when viewed as a quadratic in yy. Expanding,

(h2−ab)y2+2(hg−af)y+(g2−ac),(h^2-ab)y^2 + 2(hg-af)y + (g^2-ac),

and a quadratic Ay2+2By+CAy^2+2By+C is a perfect square exactly when its own discriminant vanishes: B2−AC=0B^2-AC=0. Here A=h2−abA=h^2-ab, B=hg−afB=hg-af, C=g2−acC=g^2-ac, so the condition is

(hg−af)2−(h2−ab)(g2−ac)=0.(hg-af)^2 - (h^2-ab)(g^2-ac) = 0.

Expanding this and simplifying (a routine but lengthy algebraic exercise) reduces it to the compact symmetric form

abc+2fgh−af2−bg2−ch2=0,abc + 2fgh - af^2 - bg^2 - ch^2 = 0, …