Question

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

Answer 1

Equation of line $\Rightarrow y=mx+c$

(1,2) is lies on this point so,

$\begin{array}{cc}2=m+c& \\ c=2-m& \\ x+y=4\&y-mx+c& \\ y=4-x\&y=mx+2-m& \\ Equatingthem,& \\ 4-x=mx+2-m& \\ 4-2+m=mx+x& \\ 2+m=x\left(m+1\right)& \\ \frac{2+m}{m+1}=x\&y=4-\left(\frac{2+m}{m+1}\right)& \\ So,P\left(\frac{2+m}{m+1},\frac{3m+2}{1+m}\right)& y=\frac{3m+2}{1+m}\\ Q\left(1,2\right)& So,\overrightarrow{PQ}=\sqrt{{\left(2-\frac{3m+2}{1+m}\right)}^2+{\left(1-\frac{2+m}{1+m}\right)}^2}\\ \overrightarrow{PQ}=\frac{\sqrt{6}}{3}& \\ Then,& \\ \frac{\sqrt{6}}{3}=\sqrt{{\left(2-\frac{3m+2}{1+m}\right)}^2+{\left(1-\frac{2+m}{1+m}\right)}^2}& \end{array}$

Square both sides & solve this

$m=2\pm \sqrt{3}$

Put the value of m in equation of line

$\begin{array}{c}y=\left(2\pm \sqrt{3}\right)x+2-2\pm \sqrt{3}\\ y=\left(2\pm \sqrt{3}\right)x\pm \sqrt{3}\end{array}$