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 then $2\sqrt{2}$. 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 intersection point of two lines is $(\frac{-c}{a+b},\frac{-c}{a+b})$
Distance between (1,1) and $(\frac{-c}{a+b},\frac{-c}{a+b})<2\sqrt{2}$
$\Rightarrow 2(1+\frac{c}{a+b}{)}^2<8\Rightarrow 1+\frac{c}{a+b}<2\Rightarrow a+b-c>0$