Question

Find the direction in which a straight line must be drawn through the point $(1,2)$, so that its point of intersection with the line $x+y=4$ may be at a distance $\frac{\sqrt{6}}{3}$ from this point.

Answer 1

$y=mx+c$

$2=m+c$

$c=2-m$

$y=mx+2-m$

I.P. with $x+y=4$ is

$x+mx+2-m=4$

$(m+1)x=2+m$

$x=\frac{m+2}{m+1}y=\frac{m^2+2m-m^2+m+3}{m+1}$

$x=\frac{m+2}{m+1}y=\frac{3m+3}{m+1}=3$

$(\frac{m+2}{m+1},3)$ at a distance of $\frac{\sqrt{6}}{3}$ from (1, 2).

$\sqrt{1^2+{(\frac{m+2}{m+1}-1)}^2}=\frac{\sqrt{6}}{3}$

$1+{\left(\frac{1}{m+1}\right)}^2=\frac{6}{9}$

$\frac{1}{m+1}=\frac{1}{\sqrt{3}}m=\sqrt{3}-1$