For $a>b>c>0$,the distance between $(1,1)$ and the point of intersection of the lines $ax+by+c=0$ and the $bx+ay+c=0$ is less than $2\sqrt2$

Question

For $a>b>c>0$,the distance between $(1,1)$ and the point of intersection of the lines $ax+by+c=0$ and the $bx+ay+c=0$ is less than $2\sqrt2$,then
$(A)a+b-c>0$

$(B)a-b+c<0$

$(C)a-b+c>0$

$(D)a+b-c<0$

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I found the point of intersection of lines $ax+by+c=0$ and the $bx+ay+c=0$.Point of intersection lies on the line $y=x$.
But after that,i could not see any way to solve it.Please help me.Thanks.

Answer 1

Aren't $A$ and $D$ the same?

Since $ax+by=bx+ay$, $(a-b)x=(a-b)y$ and $x=y={-c\over a+b}$

So ${\sqrt{2({-c\over a+b}-1)^2}< 2\sqrt2}\implies -1<{-c\over a+b}<3$

By $-1<{-c\over a+b}$ we have $a+b-c>0$ so $A$ and $D$ are correct.

And to tell the truth, if we are really given $a>b>c>0$ instead of $a>0, b>0, c>0$, we have $a+(b-c)>0$ and $(a-b)+c>0$ from the beginning, so all $A, C, D$ are correct.

Answer 2

Point of intersection of 2 given lines= [-c/(a+b),-c/(a+b)] now, distance between (1,1) and [-c/(a+b),-c/(a+b)] by distance formula is a+b+c<2a+2b or a+b-c>0 hence, A should be the option according to me.

Answer 3

There is a lot of distraction in the question. Below is a simple solution:

$a > b \implies a - b > 0$

Since $c > 0, a - b + c > 0$

Similarly, $a > c$

$a - c > 0$

Since $b > 0, a + b - c > 0$