Family of straight lines.
Equation of family of straight line passing through point of intersection of lines $L_1\,:\,a_1x\,+b_1y\,+c_1\,=0$ and $L_2\,:\,a_2x\,+b_2y\,+c_2\,=0$ is given by $L\,:\,L_1\,+ \lambda L_2\,=0$ , where $\lambda \in R$ , and $\lambda$ is a parameter as for every different $\lambda$ we get different member of family.
Hence family of straight line passing through the intersection of $L_1\,=0$ and $L_2\,=0$ is
$$ L\,:\,(a_1x\,+\,b_1y\,+\,c_1)\,+\,\lambda (a_2x\,+\,b_2y\,+\,c_2)\,=\,0$$
Hence equation of line $L\,:\,(a_1+\lambda a_2)x+(b_1+\lambda b_2)y+(c_1+\lambda c_2)\,=0$ represent equation of line passing through the intersection of lines $L_1=0$ and $L_2=0$.
NOTE :- (1). If $L_1\,+\lambda L_2\,=\,0$ is family of lines , then the point of intersection of all members of the family is called point of concurrency of this family.
(2). Let equation of sides BC , CA and AB of $\Delta ABC$ be $L_1\,:\,a_1x\,+b_1y\,+c_1\,=0\,,\,L_2\,:\,a_2x\,+b_2y\,+c_2\,=0$ and $L_3\,:\,a_3x\,+b_3y\,+c_3\,=0$ respectively
Equation of line through vertex A is $AD\,:\,L_2\,+\lambda L_3\,=0$.
- AD as median : D is mid point of BC satisfy in Ad and find $\lambda$ , Hence AD will be the required median.
- AD as altitude : Using product of slope of perpendicular lines is -1 , $m_{AD}.m_{BC}=-1.$ find $\lambda$ , Hence equation of altitude can be found.
- AD as angle bisector : Angle between lines AD , $L_2\,=0$ and AD , $L_3\,=0$ are equal , find $\lambda$. $$\vert \cfrac{m_{AD}-m_{L_2}}{1+m_{AD}m_{L_2}} \vert = \vert \cfrac{m_{L_3}-m_{AD}}{1+m_{AD}m_{L_3}} \vert $$
(3) Let $L_1\,:\,a_1x\,+b_1y\,+c_1\,=0 \,,\,L_2\,:\,a_2x\,+b_2y\,+c_2\,=0 $ and $ L_3\,:\,a_1x\,+b_1y\,+c_3\,=0 \,,\,L_4\,:\,a_2x\,+b_2y\,+c_4\,=0$ be the equation of sides AB , BC , CD and AD respectively then equation of diagonal AC is given by .
Line through vertex A is $L_1\,+\,\lambda L_4\,=0$
$\Rightarrow (a_1 + \lambda a_2)x + (b_1+\lambda b_2)y + (c_1+\lambda c_2)\,=0\,—-(1)$
Line through vertex C is $L_3\,+\,\mu L_4\,=0$
$\Rightarrow (a_1 + \mu a_2)x + (b_1+\mu b_2)y + (c_3+\mu c_4)\,=0\,—-(2)$
Since equation (1) and (2) represent the equation of diagonal AC hence
$$\Rightarrow \cfrac{a_1+\lambda a_2}{a_1+\mu a_2}= \cfrac{b_1+\lambda b_2}{a_1+\mu a_2}=\cfrac{c_1+\lambda c_4}{c_3+\mu c_2}$$
$$\Rightarrow (a_1+\lambda a_2)(b_1+\mu B_2)=(a_1+\mu a_2)(b_1+\lambda B_2)$$
$$\Rightarrow \cancel{a_1b_1}+\mu a_1b_2+\lambda a_2b_1+\cancel{\lambda \mu a_2b_2}=\cancel{a_1b_1}+\mu a_2b_1+\lambda a_1b_2+\cancel{\lambda \mu a_2b_2}$$
$$\therefore \mu (a_1b_2-a_2b_1)=(a_1b_2-a_2b_1)\lambda$$
$$\therefore \lambda = \mu \;=\, -\cfrac{l_1}{l_4}=-\cfrac{l_3}{l_2}$$
$$ $$ Equation of diagonal AC is $\Rightarrow L_1L_2-L_3L_4=0$
Similarly equation of diagonal BD is $L_1L_4-L_2L_3=0$