Discussion Forum

1)

For a>b>c>0, the distance between (1,1)  and the point of intersection of the lines ax+by+c=0  and bx+ay+c=0 is less than 

$2\sqrt{2}$  then 

Answer:

Option A

Explanation:

Concept involved Application of inequality sum and difference along with lengths of perpendicular  . For this type of questions involving inequality we should always check all options.

Situation  analysis check all inequalities according to options and use length of perpendicular from the point (x₁,y₁) to ax+by+c=0

 i.e,   $\frac{|ax_{1}+by_{1}+c|}{\sqrt{a^{2}+b^{2}}}$

   As, a>b>c>0

  a-c>0  and b>0

 $\Rightarrow$ a+b-c >0  ......(i)

  a-b >0 and c>0

 a+c-b>0  .......(ii)

 $\therefore$   (a) and (c) is correct

Also the point of intersection  for  ax+by+c=0 and bx+ay+c=0

 i.e,    $\left(\frac{-c}{a+b},\frac{-c}{a+b}\right)$

  The distance between (1,1) and 

                      $\left(\frac{-c}{a+b},\frac{-c}{a+b}\right)$

 i.e, less than  $2\sqrt{2}$ 

$\Rightarrow\sqrt{\left(1+\frac{c}{a+b}\right)^{2}+\left(1+\frac{c}{a+b}\right)^{2}}<2\sqrt{2}$

    $\Rightarrow\left(\frac{a+b+c}{a+b}\right)\sqrt{2}<2\sqrt{2}$

$\Rightarrow $   a+b+c <2a+2b

   or a+b-c >0

  From Eqs.(i) and (ii) option (c) is correct.