If a line” l ” passes through k,2k,3k,3k and 3,1 , k≠0 then the distance from the origin to the line ‘l ’ is

3k−2k3k−k=1−3k3−3k⇒k=−13l is y−1=−13x−3⇒x−2y−1=0d=15

If a line” l ” passes through k,2k,3k,3k and 3,1 , k≠0 then the distance from the origin to the line ‘l ’ is

3k−2k3k−k=1−3k3−3k⇒k=−13l is y−1=−13x−3⇒x−2y−1=0d=15

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If a line” $l$ ” passes through $(k, 2 k), (3 k, 3 k)$ and $(3, 1)$ , $k \neq 0$ then the distance from the origin to the line ‘ $l$ ’ is

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$\frac{4}{\sqrt{5}}$

$\frac{3}{\sqrt{5}}$

$\frac{2}{\sqrt{3}}$

$\frac{1}{\sqrt{5}}$

answer is D.

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Detailed Solution

$\frac{3 k - 2 k}{3 k - k} = \frac{1 - 3 k}{3 - 3 k} \Rightarrow k = \frac{- 1}{3}$

$l$ is $y - 1 = \frac{- 1}{3} (x - 3) \Rightarrow x - 2 y - 1 = 0$

$d = \frac{1}{\sqrt{5}}$

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If a line” l ” passes through k,2k,3k,3k and 3,1 , k≠0 then the distance from the origin to the line ‘l ’ is

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If a line” $l$ ” passes through $(k, 2 k), (3 k, 3 k)$ and $(3, 1)$ , $k \neq 0$ then the distance from the origin to the line ‘ $l$ ’ is