If a line l passes through (k, 2k), (3k, 3k) | vedclass

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(A) Since the points $A(k, 2k), B(3k, 3k)$,and $C(3, 1)$ are collinear,the slope of $AB$ must be equal to the slope of $BC$.Slope of $AB = \frac{3k -

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

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(A) Since the points $A(k, 2k), B(3k, 3k)$,and $C(3, 1)$ are collinear,the slope of $AB$ must be equal to the slope of $BC$.Slope of $AB = \frac{3k - 2k}{3k - k} = \frac{k}{2k} = \frac{1}{2}$.Slope of $BC = \frac{1 - 3k}{3 - 3k}$.Equating the slopes: $\frac{1}{2} = \frac{1 - 3k}{3 - 3k}$.Cross-multiplying: $3 - 3k = 2(1 - 3k)$ $\Rightarrow 3 - 3k = 2 - 6k$ $\Rightarrow 3k = -1$ $\Rightarrow k = -\frac{1}{3}$.The equation of the line passing through $B(-1, -1)$ and $C(3, 1)$ is $y - 1 = \frac{1 - (-1)}{3 - (-1)}(x - 3)$.$y - 1 = \frac{2}{4}(x - 3)$ $\Rightarrow y - 1 = \frac{1}{2}(x - 3)$ $\Rightarrow 2y - 2 = x - 3$ $\Rightarrow x - 2y - 1 = 0$.The distance from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.$d = \frac{|-1|}{\sqrt{1^2 + (-2)^2}} = \frac{1}{\sqrt{1 + 4}} = \frac{1}{\sqrt{5}}$.

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

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