For $a > b > c > 0$ , the distance between (1,1) and the point of intersection of the lines $a x + b y + c = 0$ and $b x + a y + c = 0$ is less then $2 \sqrt{2}$ . Then

a + b - c > 0

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a - b + c < 0

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a - b + c > 0

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a + b - c < 0

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Solution

The correct option is A $a + b - c > 0$
The intersection point of two lines is $(\frac{- c}{a + b}, \frac{- c}{a + b})$
Distance between (1,1) and $(\frac{- c}{a + b}, \frac{- c}{a + b}) < 2 \sqrt{2}$
$\Rightarrow 2 (1 + \frac{c}{a + b})^{2} < 8 \Rightarrow 1 + \frac{c}{a + b} < 2 \Rightarrow a + b - c > 0$

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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 then 2√2. Then

The correct option is A a+b−c>0The intersection point of two lines is (−ca+b,−ca+b) Distance between (1,1) and (−ca+b,−ca+b)<2√2 ⇒2(1+ca+b)2<8⇒1+ca+b<2⇒a+b−c>0

For $a > b > c > 0$ , the distance between (1,1) and the point of intersection of the lines $a x + b y + c = 0$ and $b x + a y + c = 0$ is less then $2 \sqrt{2}$ . Then