Pair of Straight Lines
Two lines at once — a single second-degree equation that bundles a pair of straight lines, and the algebra that pulls them back apart
- How a product of two linear factors becomes a single second-degree combined equation.
- That \(ax^2+2hxy+by^2=0\) always represents a pair of lines through the origin.
- The sum and product of the slopes, \(m_1+m_2=-\tfrac{2h}{b}\) and \(m_1m_2=\tfrac{a}{b}\).
- How \(h^2-ab\) decides whether the lines are real, coincident or imaginary, and when they are perpendicular.
- The angle between the pair and the equation of its angle bisectors.
- The condition for the general second-degree equation to be a pair, its point of intersection, and the homogenization trick.
The Combined Equation of Two Lines
A point lies on either of two lines exactly when it satisfies at least one of their equations — that is, when the product of the two left-hand sides is zero. So if the lines are \(L_1\equiv a_1x+b_1y+c_1=0\) and \(L_2\equiv a_2x+b_2y+c_2=0\), the single equation \(L_1L_2=0\) represents both at once. Multiplying out always gives a second-degree equation.
Conversely, a second-degree equation represents a pair of straight lines precisely when its left side factorises into two linear factors. The whole chapter is about recognising when that happens and reading the lines back off.
The pair \(x-2y=0\) and \(x-3y=0\) has combined equation \((x-2y)(x-3y)=x^2-5xy+6y^2=0\). Two lines, one tidy quadratic.
The Homogeneous Equation \(ax^2+2hxy+by^2=0\)
When both lines pass through the origin their constant terms vanish, and the combined equation contains only second-degree terms — it is homogeneous. Every such equation, \(ax^2+2hxy+by^2=0\), is a pair of lines through the origin.
To see why, divide by \(x^2\) and set \(m=\tfrac{y}{x}\) (the slope of a line \(y=mx\) through the origin). The equation becomes a quadratic in \(m\):
Sum & Product of the Slopes
Because \(m_1\) and \(m_2\) are the roots of \(bm^2+2hm+a=0\), Vieta's formulas hand us their sum and product immediately — without ever solving for the lines themselves.
Most quick facts about the pair follow from these two. The angle, the perpendicularity test, and conditions like "one line bisects the axes" all reduce to symmetric functions of \(m_1,m_2\).
Nature of the Lines
Whether the two lines are genuinely distinct, identical, or not real at all is decided by the discriminant of the slope quadratic, \((2h)^2-4ab=4(h^2-ab)\). The single quantity \(h^2-ab\) tells the whole story.
\(h^2>ab\): two different real lines through the origin.
\(h^2=ab\): the lines merge into one (a repeated line).
\(h^2
The lines are perpendicular exactly when \(m_1m_2=-1\), i.e. \(\tfrac{a}{b}=-1\), which rearranges to the memorable rule \(a+b=0\) — the coefficient of \(x^2\) plus the coefficient of \(y^2\) is zero.
The Angle Between the Pair
Feeding \(m_1+m_2\) and \(m_1m_2\) into the angle formula from Chapter 16 collapses everything into the coefficients. The result is the central formula of this chapter.
The lines are parallel or coincident when \(h^2=ab\) (then \(\tan\theta=0\)), and perpendicular when \(a+b=0\) (the denominator vanishes, so \(\theta=90^{\circ}\)) — consistent with Section 17-4.
The Angle Bisectors
The two lines that bisect the angles between the pair are themselves a pair through the origin — and, like all angle bisectors, they are mutually perpendicular. Their combined equation is strikingly compact.
Cross-multiplying gives \(h(x^2-y^2)=(a-b)xy\). Notice the coefficients of \(x^2\) and \(y^2\) in this equation sum to zero — the algebraic signature that the bisectors are at right angles.
The General Second-Degree Equation
When the lines do not pass through the origin, the general second-degree equation carries first-degree terms too:
When the determinant vanishes the left side factorises into two linear factors, and the equation is a pair of straight lines.
If it is a pair, the two lines cross at a single point. That intersection is the common solution of the two partial conditions \(ax+hy+g=0\) and \(hx+by+f=0\).
Lines Joining the Origin to a Curve
A favourite technique: a line cuts a second-degree curve in two points, and we want the pair of lines joining the origin to those two points — without finding the points themselves. The trick is homogenization: make the curve's equation homogeneous of degree two using the line.
For the curve \(ax^2+2hxy+by^2+2gx+2fy+c=0\) met by the line \(lx+my=1\), replace each \(1\) in the first- and zero-degree terms by \((lx+my)\). The result is homogeneous — a pair of lines through the origin — passing through both intersection points.
Putting It to Work
Problem. Find the separate lines represented by \(x^2-5xy+6y^2=0\).
Solution. Treat it as a quadratic in \(x\) and factorise the homogeneous form directly:
Problem. Find the angle between the lines \(x^2+4xy+y^2=0\).
Solution. Here \(a=1,\ b=1,\ h=2\), so \(\tan\theta=\left|\dfrac{2\sqrt{h^2-ab}}{a+b}\right|\):
Problem. Show that \(3x^2-8xy-3y^2=0\) is a pair of perpendicular lines, and find them.
Solution. Since \(a+b=3+(-3)=0\), the lines are perpendicular. Factorising confirms it:
Problem. Find the bisectors of the angles between \(x^2-5xy+6y^2=0\).
Solution. With \(a=1,\ b=6,\ h=-\tfrac52\), use \(\dfrac{x^2-y^2}{a-b}=\dfrac{xy}{h}\):
Problem. Show that \(2x^2+5xy+3y^2+6x+7y+4=0\) represents a pair of lines, and find them and their point of intersection.
Solution. Here \(a=2,\ h=\tfrac52,\ b=3,\ g=3,\ f=\tfrac72,\ c=4\). The determinant condition gives \(abc+2fgh-af^2-bg^2-ch^2=0\), so it is a pair. Factorising:
Solving \(x+y+1=0\) and \(2x+3y+4=0\) simultaneously gives the intersection \((1,-2)\).
Problem. Find the pair of lines joining the origin to the points where \(x+y=1\) meets \(x^2+y^2=1\).
Solution. Homogenize \(x^2+y^2=1\) by replacing the constant \(1\) with \((x+y)^2\):
The required lines are \(x=0\) and \(y=0\) — the coordinate axes, since the chord meets the circle at \((1,0)\) and \((0,1)\).
Chapter Summary
\(L_1L_2=0\) represents both lines; a pair exists when the quadratic factorises.
\(ax^2+2hxy+by^2=0\); slopes solve \(bm^2+2hm+a=0\).
\(m_1+m_2=-\tfrac{2h}{b}\), \(m_1m_2=\tfrac{a}{b}\).
\(h^2>ab\) distinct, \(=ab\) coincident, \(
\(\tan\theta=\left|\tfrac{2\sqrt{h^2-ab}}{a+b}\right|\); bisectors \(\tfrac{x^2-y^2}{a-b}=\tfrac{xy}{h}\).
Pair when the \(3\times3\) determinant is \(0\); homogenize a curve with a line for the lines through the origin.
Problems
Read off \(a,b,h\) first, then reach for the right tool — factor, slope relations, angle, or determinant. Difficulty rises down the list.
- Find the separate lines represented by \(x^2-9y^2=0\).
- Find the separate lines represented by \(6x^2+5xy-6y^2=0\).
- Find the combined equation of the two lines through the origin with slopes \(2\) and \(-3\).
- For what value of \(k\) does \(kx^2+4xy+y^2=0\) represent a perpendicular pair?
- For what value of \(h\) are the lines \(x^2+2hxy+9y^2=0\) coincident?
- Find the angle between the lines \(x^2+4xy+y^2=0\).
- Show that \(2x^2+7xy+3y^2=0\) represents two distinct real lines, and find them.
- Find the equation of the bisectors of the angles between \(6x^2+5xy-6y^2=0\).
- Show that \(6x^2+13xy+6y^2+7x+8y+2=0\) represents a pair of straight lines, and find them.
- Find the point of intersection of the pair in the previous problem.
- Find the condition that one of the lines of \(ax^2+2hxy+by^2=0\) is \(y=x\).
- Find the equation of the lines joining the origin to the points of intersection of \(x^2+y^2=10\) and \(x+y=4\).