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

$a+b-c>0$

As $a>b>c>0$
$a-c>0$ and $b>0$
$\Rightarrow a+b-c>0\dots \dots \left(i\right)$
$a-b>0$ and $c>0$
$a+c-b>0\dots \dots \left(ii\right)$
From Eqs. (i) and (ii), option (b) and (d) are eliminated.
Also, the point of intersection for $ax+by+c=0$ and $bx+ay+c=0$
i.e $(\frac{-c}{a+b},\frac{-c}{a+b})$
The distance between (1, 1) and $(\frac{-c}{a+b},\frac{-c}{a+b})$
i.e. less than $2\sqrt{2}$
$\Rightarrow \sqrt{{(1+\frac{c}{a+b})}^2+{(1+\frac{c}{a+b})}^2}<2\sqrt{2}\Rightarrow \left(\frac{a+b+c}{a+b}\right)\sqrt{2}<2\sqrt{2}\Rightarrow a+b+c<2a+2b\Rightarrow a+b-c>0$
$\therefore$ Option (a) is correct.