Question

Find the direction in which a straight line may be drawn through the point $(2,1)$ so that its point of intersection with the line $4y-4x+4+3\sqrt{2}+3\sqrt{10}=0$ is at a distance of $3$ unit from $(2,1)$

Answer 1

REF.Image.

Let line is drawn at a direction at

angle $\theta$ from x-axis, $\Rightarrow$ slope$=tan\theta$

coordinates of point P would be

$(2\pm 3cos\theta ,1\pm 3sin\theta )$

Taking positive signs, and putting in lines $e{q}^n$

$4(1+3sin\theta )-4(2+3cos\theta )+3\sqrt{2}+3\sqrt{10}=0$

$-4+12sin\theta -12cos\theta +3\sqrt{2}+3\sqrt{10}=0$

$\frac{1}{\sqrt{2}}(sin\theta -cos\theta )=\left(\frac{4-3\sqrt{2}-3\sqrt{10}}{12}\right)\frac{1}{\sqrt{2}}$

$\Rightarrow sin(\theta -\frac{\pi }{4})=\frac{2\sqrt{2}-3-3\sqrt{5}}{12}$

$[\theta =\frac{\pi }{4}+si{n}^{-1}(\frac{2\sqrt{2-3-3\sqrt{5}}}{12}\left)\right]$

For negative signs,

$-sin(\theta -\pi /4)=\frac{2\sqrt{2}-3-3\sqrt{5}}{12}$

$\Rightarrow [\theta =\frac{\pi }{4}+si{n}^{-1}(\frac{3+3\sqrt{5}-2\sqrt{2}}{12}\left)\right]$