Question

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

PLAN Application of inequality sum and differences, along with lengths of perpendicular. For this type of questions involving inequality we should always check all options.
Situation analysis Check all the inequalities according to options and use length of perpendicular from the point
i.e. $\frac{\left(x_1,y_1\right)toax+by+c=0}{\sqrt{a^2+b^2}}$
As $a>b>c>0$
$a-c>0$ and $b>0$
$\Rightarrow a+b-c>0\dots$ (i)
$a-b>0$ and $c>\mathrm{0....}$(ii)
$a+c-b>0$
$\therefore$ (a) and (c) are 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 $2\sqrt{2}$.
$\Rightarrow \sqrt{{\left(1+\frac{c}{a+b}\right)}^2+{\left(1+\frac{c}{a+b}\right)}^2}<2\sqrt{2}$
$\Rightarrow \left(\frac{a+b+c}{a+b}\right)\sqrt{2}<2\sqrt{2}$
$\Rightarrow a+b+c<2a+2b$
$\Rightarrow a+b-c>0$
From Eqs. (i) and (ii), option (c) is correct.