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# Pairs of Straight Lines

An analysis of the equations associated with pairs of straight lines

## Definition

Any two lines through the Origin may be written as $\inline&space;y&space;=&space;mx$ and $\inline&space;y&space;=&space;tx$ where $\inline&space;m$ and $\inline&space;t$ are their gradients. So $\inline&space;(y&space;-&space;mx)(y&space;-&space;tx)&space;=&space;0$ giving $\inline&space;y&space;-&space;mx$ or $\inline&space;y&space;-&space;tx&space;=&space;0$ must represent the pair.
The general form of this equation is given by:
$ax^2+2hxy+by^2=0$

This equation must represent a pair of straight lines, real or imaginary, through the origin. These can be written as:
$b\left(\frac{y}{x}&space;\right)^2&space;+&space;2h\left(\frac{y}{x}&space;\right)&space;+&space;a&space;=&space;0$

Since $\inline&space;\displaystyle&space;\frac{y}{x}$ is the gradient of a line through the origin
the roots of this equation must be the gradients of the lines $\inline&space;m$ and $\inline&space;t$.
Therefore $\inline&space;\displystyle&space;m&space;+&space;t=&space;-&space;\frac{2h}{t}\f$ and $\inline&space;&space;\displaystyle&space;m&space;t&space;=&space;\frac{a}{b}$

## Angles Between Lines

Suppose that the lines $\inline&space;y&space;=&space;mx$ and $\inline&space;y&space;=&space;tx$ are represented by the following equation:
$ax^2&space;+&space;2hxy&space;+&space;by^2&space;=&space;0$
If the angle between them is $\inline&space;\theta$ then:
$\tan&space;\theta&space;&space;=&space;\frac{m&space;-&space;t}{1&space;-&space;mt}=&space;\frac{\sqrt{(m&space;+&space;t)^2&space;-&space;4mt}}{1&space;+&space;mt}$
Hence
$tan\;\theta&space;&space;=&space;\frac{\sqrt{4h^2/b^2&space;-&space;4a/b}}{1&space;+&space;a/b}$
therefore
$\tan&space;\theta&space;&space;=&space;\frac{2\sqrt{h^2&space;-&space;ab}}{a&space;+&space;b}$

N.B. The lines will be parallel if the values of this fraction become infinite. i.e. $\inline&space;a&space;+&space;b&space;=&space;0$

## To Find The Equation Of The Angle Bisectors

An angle bisector divides the angle into two angles with equal measures. An angle only has one bisector. Each point of an angle bisector is equidistant from the sides of the angle.
As before suppose that the lines $\inline&space;y&space;=&space;mx$ and $\inline&space;y&space;=&space;tx$ are represented by:
$ax^2&space;+&space;2hxy&space;+&space;by^2&space;=&space;0$

The equation of the angle bisectors will be:
$\frac{y&space;-&space;mx}{\sqrt{1&space;+&space;m^2}}&space;=&space;\pm&space;\frac{y&space;-&space;tx}{\sqrt{1&space;+&space;t^2}}$
$\therefore\;\;\;\;\;(1&space;+&space;t^2)(y&space;-&space;mx)^2&space;=&space;(1&space;+&space;m^2)(y&space;-&space;tx)^2$
or
$x^2(m^2&space;-&space;t^2)&space;-&space;2xy(m&space;+&space;mt^2\;-t\;-tm^2)&space;+&space;y^2(t^2&space;-&space;m^2)&space;=&space;0$
Since $\inline&space;m$ is not equal to $\inline&space;t$, divide the above equation by $\inline&space;(m&space;-&space;t)$
$x^2(m&space;+&space;t)&space;-&space;2xy(1&space;-&space;mt)&space;-&space;y^2(m&space;+&space;t)&space;=&space;0$

Substituting for $\inline&space;(m+t)$ and $\inline&space;mt$:
$x^2(-&space;\frac{2h}{b})&space;-&space;2xy(1&space;-&space;\frac{a}{b})&space;-&space;y^2(-\frac{2h}{b})&space;=&space;0$
or
$(x^2&space;-&space;y^2)(-&space;2h)&space;=&space;2xy(b&space;-&space;a)$
Therefore the required equation is
$\frac{x^2&space;-&space;y^2}{xy}&space;=&space;\frac{a&space;-&space;b}{h}$

## To Find The Equation Of The Pair Of Lines Joining The Points Of Intersection Of The Following Two Lines, To The Origin:

$ax^2&space;+&space;2hxy&space;+&space;by^2&space;+&space;2gx&space;+&space;2fy&space;+&space;c&space;=&space;0$
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$lx&space;+&space;my&space;+&space;n&space;=&space;0$

From the linear equation express 1 as a linear function of $\inline&space;x$ and $\inline&space;y$. i.e.:
$1\;=&space;-&space;\frac{(lx&space;+&space;my)}{n}$

Use this to build up every term of the quadratic equation to the second degree and we get:
$ax^2&space;+&space;2hxy&space;+&space;by^2&space;+&space;(2gx&space;+&space;2fy)\left(-&space;\frac{lx&space;+&space;my}{n}&space;\right)&space;+&space;c\left(-&space;\frac{lx&space;+&space;my}{n}&space;\right)^2&space;=&space;0$

Every term here is of the second degree and since any point which satisfies both:
$-&space;\frac{(lx&space;+&space;my)}{n}&space;=&space;1$
and
$2hxy&space;+&space;by^2&space;+&space;2gx&space;+&space;2fy&space;+&space;c&space;=&space;0$
must also satisfy this new equation, it must represent the required pair of lines.

## To Find The Condition That The General Equation Of The Second Degree Should Represent A Pair Of Straight Lines.

So far we have considered only pairs of straight lines through the origin. The equation of the pair of lines $\inline&space;ax&space;+&space;by&space;+&space;c&space;=&space;0$ and $\inline&space;lx&space;+&space;my&space;+&space;n&space;=&space;0$ is obviously given by the equation:
$(ax&space;+&space;by&space;+&space;c)(lx&space;+&space;my&space;+&space;n)&space;=&space;0$
And it is worth noting that the equation:
$a(x&space;-&space;\alpha&space;)^2&space;+&space;2h(x&space;-&space;\alpha&space;)(y&space;-&space;\beta&space;)&space;+&space;b(y&space;-&space;\beta&space;)^2&space;=&space;0$
represents a pair of straight lines through the point $\inline&space;(\alpha,&space;\beta&space;)$ and parallel to the pair given by:
$ax^2&space;+&space;2hxy&space;+&space;by^2&space;=&space;0$

The general equation in the second degree:
$ax^2&space;+&space;2hxy&space;+&space;by^2&space;+&space;2gx&space;+&space;2fy&space;+&space;c&space;=&space;0$
will represent a pair of straight lines if it factorizes. Expanding the equation as a quadratic in x we get:
$ax^2&space;+&space;2x(hy&space;+&space;g)&space;+&space;(by^2&space;+&space;2fy\;+c)&space;=&space;0$

When we solve for $\inline&space;x$ we will get an expression containing a square root. If the equation represents a pair of lines $\inline&space;x$ must be expressible as one or other of two linear expressions in $\inline&space;x$ and $\inline&space;y$ and so this square root must be rational. $\inline&space;(hy&space;+&space;g)^2&space;-&space;a(by^2&space;+&space;2fy+c)$ must be a perfect square.
The condition for this is given by:
$(hy&space;-&space;af)^2&space;=&space;(h^2&space;-&space;ab)(g^2&space;-&space;ac)$
Which simplifies to become:
$af^2&space;+&space;bg^2&space;+&space;ch^2&space;=&space;2fgh&space;+&space;abc$
Example:
##### Example - Simple angle example
Problem
Find the Angle between the pair
$3x^2&space;-&space;4xy\;-7y^2&space;=&space;0$

Workings
The standard form for the equation is given by:
$ax^2&space;+&space;2hxy&space;+&space;by^2&space;=&space;0$
From which it can be seen that $\inline&space;a&space;=&space;3$; $\inline&space;h&space;=&space;-2$ and $\inline&space;b&space;=&space;-7$. Substituting in equation (2), then
$\tan&space;\theta&space;&space;=&space;\frac{2\sqrt{4&space;+&space;21}}{-&space;4}&space;=&space;\frac{2\times&space;5}{4}$
Solution
And the angle between the lines is $\inline&space;\tan^{_-1}\frac{5}{4}$