In this article we will try to discuss almost everything we need to know about pair of straight lines with examples. An equation is said to be the equation of pair of straight lines if the equation is the product of the equation of two straight lines is equal to zero.

Consider two equation of straight lines:

$$ax+by+c=0\quad\text{…Eqn(1)}\\dx+ey+f=0

\quad\text{…Eqn(2)}$$

So, $(ax+by+c)(dx+ey+f)=0$ is the equation of pair of straight lines.

## Condition for a equation to be the equation of pair of straight lines

Consider the * general equation of second degree*:

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

Certainly this is a non homogeneous equation. (Because Degree of all terms is not equal). This can be any equation of circles, or equation of any * conic sections*. This equation can also be the equation of pair of straight lines.

So, the equation will be the equation of pair of straight lines if the * determinant* of the equation is equal to zero.

$$\left|\begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix} \right| =0$$

* Example:* Show that $x^{2}-y^{2}-x+3y-2=0,$ represent a pair of straight line.

* Solution:* Comparing the given equation with general equation of second degree, $ax^{2}+by^{2}+2gx+2fy+2hxy+c=0,$

we get, a=1, b=-1, 2f=3, =>f=$\frac{3}{2},$ 2g=-1 =>g=-$\frac{1}{2},$ h=0, c=-2.

So, the determinant of the equation:

$$\left|\begin{matrix} a & h & g \\ h & b & f \\ g & f & c \end{matrix} \right| =

\left|\begin{matrix} 1 & 0 & -\frac{1}{2} \\ 0 & -1 & \frac{3}{2} \\ -\frac{1}{2} & \frac{3}{2} & -2 \end{matrix} \right|$$

$$=1(2-\frac{9}{4})-0(…)-\frac{1}{2}(0-\frac{1}{2})$$

$$-\frac{1}{4}+\frac{1}{4}=0$$

So, the given equation represent the equation of pair of straight lines.

## Intersection points of pair of straight lines

Suppose the equation, $ax^{2}+by^{2}+2gx+2fy+2hxy+c=0,$ represents the equation of pair of straight lines. Find the intersection point of two lines.

So to find the intersection point, we need to determine the partial derivative of the equation with respect to x, and partial derivative with respect to y.

So partial with respect to x:

$$2ax+2g+2hy=0\quad\text{…Eqn(3)} $$

And partial with respect to y:

$$2by+2f+2hx=0\quad\text{…Eqn(4)}$$

Now, solving Eqn(3), and Eqn(4) gives the value of x, and y. So, (x,y) is the intersection point of the equation.

* Example:* Find the intersection point of the lines $3x^{2}+8xy-3y^{2}+2x-4y-1=0$

**Solution:**

$$6x+8y+2=0\quad\text{partial derivative with respect to x}$$

$$8x-6y-4=0\quad\text{partial derivative with respect to y}$$

Now, solving the equations gives, $x=\frac{1}{5},$ $y=-\frac{2}{5}$. Therefore, the intersection point ( $\frac{1}{5},$ $-\frac{2}{5}$)

## Angle between lines

If the general equation of second degree, $ax^{2}+by^{2}+2gx+2fy+2hxy+c=0$ represents the equation of pair of straight lines, then the angle between two lines will be:

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

if angle, θ=0°, than

$tan 0= \frac{2\sqrt{h^{2}-ab}}{a+b}$

$=> \frac{2\sqrt{h^{2}-ab}}{a+b}=0$

$=>2\sqrt{h^{2}-ab}=0$

$=>h^{2}=ab$

So, when $h^{2}=ab$ the lines are parallel

if angle θ=90°, than

$tan90°= \frac{2\sqrt{h^{2}-ab}}{a+b}$

$=>\frac{1}{0}=\frac{2\sqrt{h^{2}-ab}}{a+b}$

$=>a+b=0$

so, this can be said that when $a+b=0,$ then the lines are perpendicular.

Again, when $h^{2}-ab=0,$ then the both lines are on the * same* line.

if $h^{2}-ab>0,$ then the lines are * real* and

*.*

**distinct**And if $h^{2}-ab<0,$ then the lines are imaginary.

## Equation of Bisector

Bisector is the line that divide the angle between two lines in equal sects.

If an equation: $ax^{2}+by^{2}+2gx+2fy+2hxy+c=0$ represents the equation of pair of straight line, and $(x_{1}, y_{1})$ represents the intersection point of two lines, then the equation of bisector will be represented by the equation:

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

## Pair of straight line passing through the origin, (0,0)

Two equations of the lines passing through the origin:

$$y=m_{1}x, =>y-m_{1}x=0\\ y=m_{2}x=>y-m_{2}x=0$$

So, the equation of pair of straight line of this lines:

$$(y-m_{1}x)(y-m_{2}x)=0\\

=>m_{1}m_{2}x^{2}-xy(m_{1}+m_{2})+y^{2}=0\quad\text{ …Eqn(5)}$$

Here, $m_{1},$ $m_{2}$ are the slopes of two lines. Equation that passes through the origin is called as Homogeneous equation. So, the above equation is a homogeneous equation. But the simple form of Homogeneous equation of second degree is represented by:

$$ax^{2}+2hxy+by^{2}=0\\

=>\frac{a}{b}x^{2}+\frac{2h}{b}xy+y^{2}\quad\text{ …Eqn(6)}$$

Comparing Eqn(5), and Eqn(6), we get, $m_{1}m_{2}=\frac{a}{b},$ and $m_{1}+m_{2}=-\frac{2h}{b}$

Therefore, the above relations represent that if the pair of straight lines passes through the origin, then the summation of their slopes is:

$$m_{1}+m_{2}=-\frac{2h}{b}$$

and multiplication of their slope is:

$$m_{1}m_{2}=\frac{a}{b}$$

//This is the end of discussion on * Pair of straight lines*.