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)

$a-b+c>0$

$ax+by+c=0\cdots \left(1\right)$
$bx+ay+c=0\cdots \left(2\right)$

$\left(1\right)\times a-\left(2\right)\times b$, we get
$x=\frac{-c}{a+b}$
Since, the two lines are symmetric about the line $y=x$, the intersection point lies on the line $y=x$.
$\therefore y=\frac{-c}{a+b}$
Or, substitute the value of $x$ in eqn $\left(1\right)$, we get $y=\frac{-c}{a+b}$

Now, $\sqrt{{(1+\frac{c}{a+b})}^2+{(1+\frac{c}{a+b})}^2}<2\sqrt{2}$

Squaring both the sides (as terms are positve)

$\Rightarrow 2{\left(\frac{a+b+c}{a+b}\right)}^2<8$

$\Rightarrow {\left(\frac{a+b+c}{a+b}\right)}^2<4$

$\Rightarrow -2<\left(\frac{a+b+c}{a+b}\right)<2$

$\Rightarrow -2a-2b<a+b+c<2a+2b$

$\Rightarrow -2a-2b<a+b+c\text{or}a+b+c<2a+2b$

$\Rightarrow -3a-3b-c<0\text{or}-a-b+c<0$

$\Rightarrow 3a+3b+c>0\text{or}a+b-c>0$

Since, $a+b-c>0$ and $a>b$
$\therefore a-b+c>0$

Hence, both $a+b-c>0$ and $a-b+c>0$ are correct.

Alternate solution:
$\because a>b>c>0$
$\Rightarrow a-b>0$ and $a-c>0$
$\Rightarrow a-b+c>0$ and $a+b-c>0$