Question

In what direction should a line be drawn through the point $(1,2)$ so that its points of intersection with the line $x+y=4$ is at a distance $\frac{\sqrt{6}}{3}$ from the given point?

Answer 1

Let slope of the lines be no. Also, the line passes through the point $A(1,2)$

$\therefore$ equation of line is $y-2=m(x-1)$

$mx-y+2-m=\mathrm{0...}\left(1\right)$

Also, the equation of the given line is $x+y-4=\mathrm{0...}\left(2\right)$

Let these lines meet at point $B$

Solving $\left(1\right)$ & $\left(2\right)$ we get

$B\equiv (\frac{m+2}{m+1},\frac{3m+2}{m+1})$

Now, given thta $AB=\frac{\sqrt{6}}{3}$

$\Rightarrow A{B}^2=\frac{6}{9}\Rightarrow {(\frac{m+2}{m+1}-1)}^2+{(\frac{3m+2}{m+1}-2)}^2=\frac{6}{9}$

$\Rightarrow {\left(\frac{1}{m+1}\right)}^2+{\left(\frac{m}{m+1}\right)}^2=\frac{2}{3}\Rightarrow \frac{1+m^2}{(1+m{)}^2}=\frac{2}{3}$

$3m^2+3=2+2m^2+4m$

$\Rightarrow m^2-4m+1=0$

$\therefore m=\frac{4\pm \sqrt{16-4}}{2}=2\pm \sqrt{3}$

$\therefore \tan \theta =2+\sqrt{3}$ or $2-\sqrt{3}$

$\theta ={75}^{o}or\theta ={15}^o$