CodeCogs

An analysis of the equations associated with pairs of straight lines

You're viewing an older version of this page (#3385). View the current version.

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

$ax^2+2hxy+by^2=0$

(1)

This equation must represent a pair of straight lines, real or imaginary, through the origin. These can be written as:

$b\left(\frac{y}{x} \right)^2 + 2h\left(\frac{y}{x} \right) + a = 0$

(2)

Since $\frac{y}{x}$ is the gradient of a line through the origin the roots of this equation must be the gradients of the lines $m$ and $t$ .

$\therefore\;\;\;\;\;m + t= - \frac{2h}{t}\f] and \[m t = \frac{a}{b}$

(3)

The Angles Between The Lines $ax^2+2hxy+by^2=0$

Suppose that the lines y = mx and y = tx are represented by te following equation:

$ax^2 + 2hxy + by^2 = 0$

(4)

If the angle between them is $\theta$ then:

$\tan \theta = \frac{m - t}{1 - mt}$

(5)

$= \frac{\sqrt{(m + t)^2 - 4mt}}{1 + mt}$

(6)

Using equations ( ) and ( )

$tan\;\theta = \frac{\sqrt{4h^2/b^2 - 4a/b}}{1 + a/b}$

(7)

therefore

$\tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b}$

(8)

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

To find the Equation of the Angle Bisectors

As before suppose that the lines y = mx and y = tx are represented by:

$ax^2 + 2hxy + by^2 = 0$

(9)

The equation of the angle bisectors will be:

$\frac{y - mx}{\sqrt{1 + m^2}} = \pm \frac{y - tx}{\sqrt{1 + t^2}}$

(10)

$\therefore\;\;\;\;\;(1 + t^2)(y - mx)^2 = (1 + m^2)(y - tx)^2$

(11)

$x^2(m^2 - t^2) - 2xy(m + mt^2\;-t\;-tm^2) + y^2(t^2 - m^2) = 0$

(12)

Since $m$ is not equal to $t$ , divide the above equation by $(m - t)$

$x^2(m + t) - 2xy(1 - mt) - y^2(m + t) = 0$

(13)

Substituting for $(m+t)$ and $mt$ :

$x^2(- \frac{2h}{b}) - 2xy(1 - \frac{a}{b}) - y^2(-\frac{2h}{b}) = 0$

(14)

$(x^2 - y^2)(- 2h) = 2xy(b - a)$

(15)

Therefore the requires equation is

$\frac{x^2 - y^2}{xy} = \frac{a - b}{h}$

(16)

To Find the Equation of the Pair of Lines joining the Points of Intersection of the following two lines, to the Origin:

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$

(17)

</p> <p>

$lx + my + n = 0$

(18)

From the linear equation express 1 as a linear function of x and y. i.e.:

$1\;= - \frac{(lx + my)}{n}$

(19)

Use this to build up every term of the quadratic equation to the second degree and we get:

$ax^2 + 2hxy + by^2 + (2gx + 2fy)\left(- \frac{lx + my}{n} \right) + c\left(- \frac{lx + my}{n} \right)^2 = 0$

(20)

Every term here is of the second degree and since any point which satisfies both:

$- \frac{(lx + my)}{n} = 1$

(21)

and

$2hxy + by^2 + 2gx + 2fy + c = 0$

(22)

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 $ax + by + c = 0$ and $lx + my + n = 0$ is obviously given by the equation:

$(ax + by + c)(lx + my + n) = 0$

(23)

And it is worth noting that the equation:

$a(x - \alpha )^2 + 2h(x - \alpha )(y - \beta ) + b(y - \beta )^2 = 0$

(24)

represents a pair of straight lines through the point $(\alpha, \beta )$ and parallel to the pair given by:

$ax^2 + 2hxy + by^2 = 0$

(25)

The general equation in the second degree:

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$

(26)

will represent a pair of straight lines if it factorizes. Expanding the equation as a quadratic in x we get:

$ax^2 + 2x(hy + g) + (by^2 + 2fy\;+c) = 0$

(27)

When we solve for $x$ we will get an expression containing a square root. If the equation represents a pair of lines $x$ must be expressible as one or other of two linear expressions in $x$ and $y$ and so this square root must be rational. $(hy + g)^2 - a(by^2 + 2fy+c)$ must be a perfect square. The condition for this is given by: