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I was trying to understand the solution to this question here when I had the following questions (at the bottom).

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

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

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

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

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

The solution :

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.

  1. I can understand how (A) and (C) are proved but fail to understand how (D) is arrived at. To me (A) and (C) seem to be contradictory. Or is it that the inequality indicates that $a+b-c \neq 0$?

(Apologies for the trivial question but I do not have the required reputation to ask in the comments)

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    $\begingroup$ How can (A) and (D) be both correct? $a+b-c$ cannot be strictly positive and strictly negative at the same time... $\endgroup$ Commented Mar 26, 2019 at 12:21
  • $\begingroup$ Yes it seems only (A) and (C) can be proved. Should I delete this post in that case? I don't have the reputation to edit the original though. $\endgroup$
    – idunno
    Commented Mar 26, 2019 at 12:36

1 Answer 1

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The intersection point is effectively $(-\frac{c}{a+b}, - \frac{c}{a+b})$, and the distance to $(1,1)$ is given by $$\sqrt{2\left(1+\frac{c}{a+b}\right)^2}= \sqrt{2}\left(1+\frac{c}{a+b}\right)$$ (note that $1+c/(a+b)>1$)

Since the distance is less than $2 \sqrt{2}$, we know that $$ 1+ \frac{c}{a+b} < 2 \Leftrightarrow \frac{c-a-b}{a+b} < 0 \Leftrightarrow c-a-b < 0 \Leftrightarrow a+b-c >0 $$ (note that $a+b>0$)

This corresponds to option (A).

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