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Family of Straight lines. - Best Maths Practice Material

Family of Straight lines.

 

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.   

                                  Family of lines

         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   $AD\,:\,L_2\,+\lambda L_3\,=0$.

line passing through vertex A of triangle ABC

  • 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   $L_1L_4-L_2L_3=0$