Pair of straight line.

Pair of lines

Equation of pair of straight line.

                    What does concept of pair of straight line represent.

                    Let $L_1\,:\,a_1x+b_1y+c_1=0$ and  $L_2\,:\,a_2x+b_2y+c_2=0$ be two lines, then pair of straight line (also called as combined equation of given lines) is

                           $ L_1L_2\;:\;(a_1x+b_1y+c_1)(a_2x+b_2y+c_2)=0 $

                     Vise versa if pair of straight line is $L_1=0\,,\,L_2=0$ is  $(a_1x+b_1y+c_1)(a_2x+b_2y+c_2)=0$ then seperate equation of lines $L_1\,:\,a_1x+b_1y+c_1=0$ and  $L_2\,:\,a_2x+b_2y+c_2=0$.

Note :-        (1)  Equation of pair of straight lines $L_1.L_2=0$ satisfy every point that exist on either of the lines, i.e either $L_1=0$ or $L_2=0$.

                    (2)  Combined equation of pair of staight lines $L_1.L_2=0$ can be expressed as $ax^2+2hxy+by^2+2gx+2fy+c=0$ which is second degree equation in two variable.

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Condition for general second degree equation to represent pair of lines. 

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                    The second degree general equation $f(x,y)=ax^2+2hxy+by^2+2gx+2fy+c=0$ represents pair of straight lines $L_1.L_2=0$ if and only if $f(x,y)=0$ can be factorized as two linear equations.

                 i.e if we can express $f(x,y)=(a_1x+b_1y+c_1)(a_2x+b_2y+c_2)=0$  

                  Considering general second degree equation $f(x,y)=0$ as quadratic in variable $y$.

                $\Rightarrow by^2+2(hx+f)y+(ax^2+2gx+c) = 0\,—-\,(1)$  

For $f(x,y)=0$ to be factorized into two linear factors, equation (1) should have real roots $\forall x\in R$.

                 $\Rightarrow D=4(hx+f)^2-4b(ax^2+2gx+c)\ge 0\;\;\forall x\in R$  

                 Let $g(x)=(h^2-ab)x^2+2(hf-bg)x+(f^2-bc)\ge 0\;\;\forall x\in R$

                $\therefore y=\cfrac{-2(hx+f)\pm \sqrt{g(x)}}{2b}$

                 $\Rightarrow$ g(x) should be a perfect squre $\forall x\in R$ and $(h^2-ab)\gt 0$ 

                 $\Rightarrow h^2-ab\gt 0$ and Discriminant $D_g = 0$.

                 $\Rightarrow h^2-ab\gt 0\;and\; 4(hf-bg)^2-4(h^2-ab)(f^2-bc)=0$

                 $\Rightarrow \cancel{h^2f^2}+b^2g^2-2hfbg-\cancel{h^2f^2}+h^2bc+abf^2-ab^2c=0$

                  $\Rightarrow b(bg^2+ch^2+af^2-2hfg-abc)=0$

                  $\Rightarrow bg^2+ch^2+af^2-2hfg-abc=0$

                  Can also be written as $\Delta=\begin{vmatrix} a&h&g \\ h&b&f \\ g&f&c \end{vmatrix}=0$

                   Now I can conclude condition for $f(x,y)=ax^2+2hxy+by^2+2gx+2fy+c=0$  to represent pair of lines is $h^2\gt ab$  and  $\begin{vmatrix} a&h&g \\ h&b&f \\ g&f&c \end{vmatrix}=0$.

 Try It Yourself :-    Find point of intersection of line represented by pair of line $ax^2+2hxy+by^2+2gx+2fy+c=0$   

Comming Soon – Homogeneous equation of pairs of straight lines.           

Summary
Article Name
Pair Of Straight Lines
Description
Discussion on how to find pairs of lines.
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