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 $\frac{1}{3}\sqrt{6}$ from this point.

Answer 1

Let the slope of line passing $(1,2)$ be $m$

Then equation of line is

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

Given line is $x+y=4$

i.e. $y=4-x$

Substituting $y$ in $\left(i\right)$, we get

$mx-4+x+2-m=0(m+1)x=2+mx=\frac{m+2}{m+1}$

Now, $y=4-x$

$\Rightarrow y=4-\frac{m+2}{m+1}\Rightarrow y=\frac{3m+2}{m+1}$

So the point of intersection is $P(\frac{m+2}{m+1},\frac{3m+2}{m+1})$

Distance of $P$ from $(1,2)$ is $\frac{\sqrt{6}}{3}$

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

Squaring both sides, we get

${(\frac{m+2}{m+1}-1)}^2+{(\frac{3m+2}{m+1}-2)}^2=\frac{6}{9}\frac{1}{{(m+1)}^2}+\frac{m^2}{{(m+1)}^2}=\frac{2}{3}3+3m^2=2m^2+2+4m{m}^2-4m+1=0m=\frac{4\pm \sqrt{16-4}}{2}m=\frac{4\pm 2\sqrt{3}}{2}m=2\pm \sqrt{3}$

$\tan \theta =2\pm \sqrt{3}$

When $\tan \theta =2+\sqrt{3}\Rightarrow \theta ={75}^{\circ }$

When $\tan \theta =2-\sqrt{3}\Rightarrow \theta ={15}^{\circ }$

So the straight line may be inclined at angle ${75}^{\circ }$ or ${15}^{\circ }$.