Pairs of straight lines

General equation of second degree

The general equation of first degree in x and y always represents a straight line.

Let A₁x + B₁y + C₁ = 0.......................(i)

and A₂x + B₂y + C₂ = 0.......................(ii)

be the equations of two straight lines.

Now, Combining these equations we get,

(A₁x + B₁y + C₁) (A₂x + B₂y + C₂) = 0........................(iii)

The coordinates of any point, which satisfy the equation (iii), will also satisfy either equation (i) or equation (ii).

Similarly, the co-ordinates of any point, which satisfy any one of the equation (i) or (ii) will also satisfy the equation (iii).

Therefore, equation (iii) represents two separate straight lines (i) and (ii). In other words, equation (iii) represents a pair of the straight lines given by (i) and (ii). So, (iii) is the equation of a pair of lines.

Now, Expanding the left hand side of equation (iii) we get,

A₁A₂x² + (A₁B₂ + A₂B₁)xy + B₁B₂y² + (A₁C₂ + A₂C₁)x + (B₁C₂ + B₂C₁)y + C₁C₂ = 0

If we put A₁A₂ = a,A₁B₂ + A₂B₁ = 2h, B₁B₂ = b, A₁C₂ + A₂C₁ = 2g, B₁C₂ + B₂C₁ = 2f and C₁C₂ = c, then the above equation becomes,

ax² + 2hxy + by² + 2gx + 2fy + c = 0..........................(iv)

This equation is called the general equation of the second degree in x and y. Thus we see that the equation of a pair of lines is a second degree equation. But the converse of this statement is not always true. It means every second degree equations in x and y may not represent a pair of straight lines. Equations of second degree will represent a pair of straight lines only if the left hand side can be resolved into two linear factors.

Consider an equation y² - 3xy + 2x² = 0. This equation is equivalent to (y - x) (y - 2x) = 0.

So, the equation y² - 3xy + 2x² = 0 represents the two straight lines y - x = 0 and y - 2x = 0.

Similarly, the equation xy = 0 represents the two straight lines x = 0 and y = 0.

Again,

Consider an equation x² - 5x + 6 = 0 represents two straight lines x - 2 = 0 and x - 3 = 0.

And the equation x² - y² = 0 represents the two straight lines x + y = 0 and x - y = 0.

Condition that the general equation of second degree may represent a line pair

The general equation of second degree in x and y is

ax² + 2hxy + by² + 2gx + 2fy + c = 0

or, ax² + (2hy + 2g)x + (by² + 2fy + c) = 0

This is quadratic equation in x.

So, x = \(\frac {-(2hy + 2g) ± \sqrt {(2hy + 2g)^2 - 4a(by^2 + 2fy + c)}}{2a}\)

or, x = \(\frac {-(hy + g) ± \sqrt {(hy + g)^2 - a(by^2 + 2fy + c)}}{a}\)

These two equations will be linear if

(hy + g)² - a (by² + 2fy + c) is a perfect square.

or, h²y² + 2ghy + g² - aby² - 2afy - ac is a perfect square.

i.e. (h² - ab) y² + (2gh - 2af) y + (g² - ac) is a perfect square.

i.e. (2gh - 2af)² - 4 (h² - ab) (g² - ac) = 0

i.e. g²h² - 2ghaf + a²f² - g²h² + h²ac + abg² - a²bc = 0

i.e. a (af² + bg² + ch² - 2fgh - abc) = 0

i.e. abc + 2fgh - af² - bg² - ch² = 0

Hence, the general equation of second degree ax² + 2hxy + by² + 2gx + 2fy + c = 0 may represent a line pair if abc + 2fgh - af² - bg² - ch² = 0.

Quadratic Equation

Any equation in the form of ax² + bx + c = 0 is called quadratic equation in x.

Multiplying both sides of this equation by 4a we get,

4a (ax² + bx + c) = 0

or, 4a²x² + 4abx + 4ac = 0

or, (2ax)² + 2 . 2ax . b + b² - b² + 4ac = 0

or, (2ax + b)² = b² - 4ac

or, (2ax + b)² = (\(\sqrt {b^2 - 4ac}\))²

or, 2ax + b = ± \(\sqrt {b^2 - 4ac}\)

or, 2ax = - b ± \(\sqrt {b^2 - 4ac}\)

or, x = \(\frac {- b ± \sqrt {b^2 - 4ac}}{2a}\)

Let α = \(\frac {- b + \sqrt {b^2 - 4ac}}{2a}\) and β = \(\frac {- b - \sqrt {b^2 - 4ac}}{2a}\).

Then α and β are called roots of the quadratic equation ax² + bx + c = 0.

Now,

\begin{align*} \text{Sum of the roots (α + β)} &= \frac {- b + \sqrt {b^2 - 4ac}}{2a} + \frac {- b - \sqrt {b^2 - 4ac}}{2a}\\ &= - \frac ba\\ &= -\frac {coefficient\;of\;x}{coefficient\;of\;x^2}\\ \end{align*}

\begin{align*} \text{Sum of the roots (αβ)} &= (\frac {- b + \sqrt {b^2 - 4ac}}{2a}) (\frac {- b - \sqrt {b^2 - 4ac}}{2a})\\ &= \frac ca\\ &= \frac {constant\;term}{coefficient\;of\;x^2}\\ \end{align*}

Homogeneous equation of second degree

An equation in x and y in which the sum of the power of x and y in every term is the same, is called homogenous equation. If this sum is two, then the equation is called a homogenous equation of second degree. The equation ax² + 2hxy + by² = 0 is the general homogenous equation of the second degree.

A homogeneous equation of the second degree represents a pair of straight lines which pass through the origin.

Proof: Consider the homogenous equation of the second degree.

ax² + 2hxy + by² = 0...........................(i)

or, by² + 2hxy + ax² = 0

If b ≠ 0, the equation can be written as y² + \(\frac {2h}b\)xy + \(\frac ab\)x² = 0

or, (\(\frac yx\))² + \(\frac {2h}b\)(\(\frac yx\)) + \(\frac ab\) = 0

This is quadratic equation in \(\frac yx\). So it has two roots. Let these roots be m₁ and m₂.

Then,

\(\frac yx\) = m₁ and \(\frac yx\) = m₂

or, y = m₁x and y = m₂x.

These two equations are the equations of straight lines passing through the origin.

If b = 0, then equation (i) becomes

ax² + 2hxy = 0

or, x(ax + 2hy) = 0 which represents two straight lines x = 0 and ax + 2hy = 0.

These two lines pass through the origin.

Hence, the homogenous equation of second degree always represents a pair of straight lines passing through the origin.

Angle between the line pair represented by ax² + 2hxy + by² = 0

Homogeneous equation of second degree is:

ax² + 2hxy + by² = 0

or, by² + 2hxy + ax² = 0

or, y² + \(\frac{2h}{b}\)(\(\frac yx\)) + \(\frac ab\) = 0.........................(i)

This is quadratic equation in \(\frac yx\). So, it has two roots. Let these two roots be m₁ and m₂.

Then,

\(\frac yx\) = m₁ and \(\frac yx\) = m₂

or, y = m₁x and y = m₂x which are two seperate equations represente by the given equation ax² + 2hxy + by² = 0.

Now,

From the quadratic equation (i)

m₁ + m₂ = \(\frac {-2h}b\) and m₁m₂ = \(\frac ab\)

Now,

\begin{align*} m_1- m_2 &= \sqrt {(m_1 + m_2)^2 - 4m_1m_2}\\ &= \sqrt {\frac{4h^2}{b^2} - 4\frac ab}\\ &= \sqrt {\frac {4h^2 - 4ab}{b^2}}\\ &= \frac 2b \sqrt {h^2 - ab}\\ \end{align*}

Let \(\theta\) be the angle between the lines y = m₁x and y = m₂x. Then,

\begin{align*} tan\theta &= ± \frac {m_1 - m_2}{1 + m_1m_2}\\ &= ± \frac {\frac 2b \sqrt {h^2 - ab}}{1 + \frac ab}\\ &= ± \frac {2\sqrt {h^2 - ab}}{a + b}\\ \end{align*}

∴ \(\theta\) = tan^-1 (± \(\frac {2\sqrt {h^2 - ab}}{a + b}\))

Second Method:

We have ax² + 2hxy + by² = 0............................................(i)

Let the seperate equations represented by this equation be y = m₁x and y = m₂x.

Now,

The combined equation of theses equation is:

(y - m₁x) (y - m₂x) = 0

or, y² - (m₁ + m₂)xy + m₁m₂x² = 0

or, m₁m₂x²- (m₁ + m₂)xy + y² = 0.......................................(ii)

Comparing (i) and (ii) we have,

\(\frac {m_1m_2}{a}\) = \(\frac {- (m_1 + m_2)}{2h}\) = \(\frac 1b\)

Taking 1^st and last, m₁m₂ = \(\frac ab\)

Taking 2^nd and last, m₁ + m₂ = -\(\frac {2h}b\)

Now,

\begin{align*} m_1 - m_2 &= \sqrt {(m_1 + m_2)^2 - 4m_1m_2}\\ &= \sqrt {\frac {4h^2}{b^2} - 4\frac ab}\\ &= \sqrt {\frac {4h^2 - 4ab}{b^2}}\\ &= \frac 2b \sqrt {h^2 - ab}\\ \end{align*}

Let \(\theta\) be the angle between the lines y = m₁x and y = m₂x.

Then,

\begin{align*} tan\theta &=± \frac {m_1 - m_2}{1 + m_1m_2}\\ &= ± \frac {\frac 2b \sqrt {h^2 - ab}}{1 + \frac ab}\\ &= ± \frac {2\sqrt {h^2 - ab}}{a + b}\\ \end{align*}

∴ \(\theta\) = tan^-1 (± \(\frac {2\sqrt {h^2 - ab}}{a + b}\))

Condition that the straight lines given by the equation ax² + 2hxy + by² = 0 may be (1) Perpendicular and (2) Coincident.

1. If a + b = 0 the value of tan\(\theta\) is∞ and hence \(\theta\) is 90°. Hence two straight lines represented by the equation ax² + 2hxy + by² = 0 are perpendicular to each other if coefficient of x² + coefficient of y² = 0.
   For example: the equations x² - y² = 0 and 6x² + 11xy - 6y²= 0 both represent pairs of straight lines at right angles.
   Similarly, whatever be the value of h, the equation x² + 2hxy - y² = 0 represents a pair of straight lines at right angles.
2. If h² = ab, the value of tan\(\theta\) is zero and hence \(\theta\) is zero. But both the straight lines represented by ax² + 2hxy + by² = 0 pass through the origin. So, the two lines are coincident.
   Hence, two straight lines represented by the equation ax² + 2hxy + by² = 0 are coincident ifh² = ab.
   This may be seen directly from the original equation ax² + 2hxy + by² = 0. If h² = ab i.e. h = \(\sqrt {ab}\), then the original equation becomes,
   ax² + 2\(\sqrt {ab}\)xy + by² = 0
   or, (\(\sqrt a\)x)² + 2 \(\sqrt a\) \(\sqrt b\) xy + (\(\sqrt b\)y)² = 0
   or, (\(\sqrt a\)x + \(\sqrt b\)y)² = 0 which gives two coincident straight lines
   i.e.\(\sqrt a\)x + \(\sqrt b\)y = 0 and\(\sqrt a\)x + \(\sqrt b\)y = 0.

If the equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines, then ax² + 2hxy + by² = 0 represents a pair of lines through the origin parallel to the above pair.

Proof:

Let ax²+ 2hxy + by² + 2gx + 2fy + c = 0............................................................(i)

represent a pair of straight lines. Then the left-hand side can be resolved into two linear factors. Let these factors be A₁x + B₁y + C₁and A₂x + B₂y + C₂ = 0.

Now

Combining equation of these equations is:
(A₁x + B₁y + C₁) (A₂x + B₂y + C₂) = 0

or, A₁A₂x² + (A₁B₂ + A₂B₁)xy + B₁B₂y² + (A₁C₂ + A₂C₁)x + (B₁C₂ + B₂C₁)y + C₁C₂ = 0..............................................(ii)

Equating the coefficients of like terms in equation (i) and (ii) we have,

A₁A₂ = a², A₁B₂ + A₂B₁ = 2h, B₁B₂ = b², A₁C₂ + A₂C₁ = 2g, B₁C₂ + B₂C₁ = 2f, C₁C₂ = c²

Now,

Equation of the straight line parallel to A₁x + B₁y + C₁ = 0 and passing through the origin is:

A₁x+ B₁y = 0.........................................(iii)

Again,

Equation of straight line parallel to A₂x + B₂y + C₂ = 0 and passing through the origin is:

A₂x + B₂y = 0.........................................(iv)

Now,

Combining (iii) and (iv) we have,

(A₁x + B₁y) (A₂x + B₂y) = 0

or, A₁A₂x² + (A₁B₂ + A₂B₁)xy + B₁B₂y² = 0

or, ax² + 2hxy + by² = 0. This completes the proof.

Note:Angles between the line pair ax² + 2hxy + by² + 2gx + 2fy + c = 0 are same as the angles between the pair ax² + 2hxy + by² = 0.

The angles are given by,

tan^-1 (± \(\frac {2\sqrt {h^2 - ab}}{a + b}\))

Hence, the two lines will be perpendicular to each other if a + b = 0 and they will beparallel if h²= ab.