Question

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

• A. $\frac{1}{\sqrt{5}}$
• B. $\frac{4}{\sqrt{5}}$
• C. $\frac{3}{\sqrt{5}}$
• D. $\frac{2}{\sqrt{5}}$

Answer 1

Answer: (A)

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

Answer 2

Given points $A(k, 2 k), B(3 k, 3 k)$ and $C(3,1)$ are collinear. $\therefore$ Slope of $A B=$ Slope of $B C$ $\therefore \frac{2 k-2 k}{3 k-k}=\frac{1-3 k}{3-3 k}$ $\Rightarrow \frac{k}{2 k}=\frac{1-3 k}{3-3 k}$ $\Rightarrow 3-3 k=2(1-3 k)$ $\Rightarrow 1=-3 k$ $\Rightarrow k=-\frac{1}{3}$ $\therefore$ Given points become $A\left(-\frac{1}{3},-\frac{2}{3}\right), B(-1,-1)$ and $C(3,1)$. $\therefore$ Equation of line passing through $B$ and $C$ is $y+1=\frac{1+1}{3+1}(x+1)$ $\Rightarrow y+1 =\frac{2}{4}(x+1)$ $\Rightarrow 2(y+1)= (x+1)$ $\Rightarrow 2 y-x+1=0$ Now, the distance from $(0,0)$ to the above line is $d=\frac{|2(0)-0+1|}{\sqrt{2^{2}+1^{2}}} \Rightarrow=\frac{1}{\sqrt{4+1}}=\frac{1}{\sqrt{5}}$ Equation of line passing through $(k, 2 k)$ and $(3 k, 3 k)$ is $(y-2 k)=\frac{3 k-2 k}{3 k-k}(x-k)$ $\Rightarrow y-2 k=\frac{k}{2 k}(x-k)$ $\Rightarrow y-2 k=\frac{1}{2}(x-k)\,\,\,...(i)$ Since, above line is passing through $(1,1)$. $\therefore 1-2 k=\frac{1}{2}(1-k)$ $\Rightarrow 2-4 k=1-k$ $\Rightarrow 1=3 k$ $\Rightarrow k=\frac{1}{3}$ On putting $k=\frac{1}{3}$ in E (i), we get $y-\frac{2}{3}=\frac{1}{2}\left(x-\frac{1}{3}\right)$ $\Rightarrow 3 y-2=\frac{1}{2}(3 x-1)$ $\Rightarrow 6 y-4=3 x-1$ $\Rightarrow 6 y-3 x-3=0$ or $2 y-x-1=0$ $\therefore$ Perpendicular distance from $(0,0)$ to the above line is $d =\frac{|2(0)-0-1|}{\sqrt{2^{2}+1^{2}}}=\frac{1}{\sqrt{4+1}}$ $=\frac{1}{\sqrt{5}}$