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 than 2 2 . Then [2013]

Question

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 than $2 \sqrt{2}$ . Then [2013]

Smart Achievers · Accepted Answer

(1)

The intersection point of two given lines is $(\frac{- c}{a + b}, \frac{- c}{a + b})$

$Now, distance between (1, 1) and (\frac{- c}{a + b}, \frac{- c}{a + b}) < 2 \sqrt{2}$

$\Rightarrow 2 {(1 + \frac{c}{a + b})}^{2} < 8 \Rightarrow 1 + \frac{c}{a + b} < 2$

$\Rightarrow a + b - c > 0$

Solution

Q.
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 than $2 \sqrt{2}$ . Then [2013]

1 $a + b - c > 0$

2 $a - b + c < 0$

3 $a - b + c > 0$

4 $a + b - c < 0$

Ans.
(1)

The intersection point of two given lines is $(\frac{- c}{a + b}, \frac{- c}{a + b})$

$Now, distance between (1, 1) and (\frac{- c}{a + b}, \frac{- c}{a + b}) < 2 \sqrt{2}$

$\Rightarrow 2 {(1 + \frac{c}{a + b})}^{2} < 8 \Rightarrow 1 + \frac{c}{a + b} < 2$

$\Rightarrow a + b - c > 0$

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Solution

Q. 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 22. Then [2013]

1 a+b-c>0

2 a-b+c<0

3 a-b+c>0

4 a+b-c<0

Ans. (1) The intersection point of two given lines is (-ca+b,-ca+b) Now, distance between (1,1) and (-ca+b,-ca+b)<22 ⇒2(1+ca+b)2<8⇒1+ca+b<2 ⇒a+b-c>0