Question

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

• A. $a+b-c>0$
• B. $a-b+c<0$
• C. $a-b+c>0$
• D. $a+b-c<0$

Answer 1

The correct option is A $a+b-c>0$

The point of intersection for $ax+by+c=0$ and $bx+ay+c=0$

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

The distance between $(1,1)$ is $(\frac{-c}{a+b},\frac{-c}{a+b})$ is less than $2\sqrt{2}$

$\Rightarrow \sqrt{{(1+\frac{c}{a+b})}^2+{(1+\frac{c}{a+b})}^2}<2\sqrt{2}\Rightarrow \left(\frac{a+b+c}{a+b}\right)\sqrt{2}<2\sqrt{2}\Rightarrow a+b+c<2a+2b\Rightarrow a+b-c>0$

Therefore $a+b-c>0$ if the required distance is less than $2\sqrt{2}.$