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

Answer: (A), (C)

The correct options are
A

a + b - c > 0

C

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 $c>0$
$a+c-b>0\dots \dots \left(ii\right)$
$\therefore$ (a) and (c) are correct.
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 $2\sqrt{2}$
$\Rightarrow \sqrt{{(1+\frac{c}{a+b})}^2+{(1+\frac{c}{a+b})}^2}<2\sqrt{2}\Rightarrow \left(\frac{a+b+c}{a+b}\right)\sqrt{2}<2\sqrt{2}\Rightarrow a+b+c<2a+2b\Rightarrow a+b-c>0$
From Eqs. (i) and (ii), option (c) is correct.