Question
Slope of a line passing through $$P\left( {2,\,3} \right)$$ and intersecting the line $$x+y=7$$ at a distance of 4 units from $$P,$$ is :
A.
$$\frac{{1 - \sqrt 5 }}{{1 + \sqrt 5 }}$$
B.
$$\frac{{1 - \sqrt 7 }}{{1 + \sqrt 7 }}$$
C.
$$\frac{{\sqrt 7 - 1}}{{\sqrt 7 + 1}}$$
D.
$$\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}$$
Answer :
$$\frac{{1 - \sqrt 7 }}{{1 + \sqrt 7 }}$$
Solution :
Since point at 4 units from $$P\left( {2,\,3} \right)$$ will be $$A\left( {4\,\cos \,\theta + 2,\,4\,\sin \,\theta + 3} \right)$$ and this point will satisfy the equation of line $$x+y=7$$
$$ \Rightarrow \cos \,\theta + \sin \,\theta = \frac{1}{2}$$
On squaring
$$\eqalign{ & \Rightarrow \sin \,2\theta - \frac{3}{4} \Rightarrow \frac{{2\,\tan \,\theta }}{{1 + {{\tan }^2}\theta }} = - \frac{3}{4} \cr & \Rightarrow 3\,{\tan ^2}\theta + 8\,\tan \,\theta + 3 = 0 \cr & \Rightarrow \tan \,\theta = \frac{{ - 8 \pm 2\sqrt 7 }}{6}\,\,\,\,\,\,\left( {{\text{ignoring }} - {\text{ve sign}}} \right) \cr & \Rightarrow \tan \,\theta = \frac{{ - 8 + 2\sqrt 7 }}{6} \cr & \Rightarrow \tan \,\theta = \frac{{1 - \sqrt 7 }}{{1 + \sqrt 7 }} \cr} $$
Since point at 4 units from $$P\left( {2,\,3} \right)$$ will be $$A\left( {4\,\cos \,\theta + 2,\,4\,\sin \,\theta + 3} \right)$$ and this point will satisfy the equation of line $$x+y=7$$
$$ \Rightarrow \cos \,\theta + \sin \,\theta = \frac{1}{2}$$
On squaring
$$\eqalign{ & \Rightarrow \sin \,2\theta - \frac{3}{4} \Rightarrow \frac{{2\,\tan \,\theta }}{{1 + {{\tan }^2}\theta }} = - \frac{3}{4} \cr & \Rightarrow 3\,{\tan ^2}\theta + 8\,\tan \,\theta + 3 = 0 \cr & \Rightarrow \tan \,\theta = \frac{{ - 8 \pm 2\sqrt 7 }}{6}\,\,\,\,\,\,\left( {{\text{ignoring }} - {\text{ve sign}}} \right) \cr & \Rightarrow \tan \,\theta = \frac{{ - 8 + 2\sqrt 7 }}{6} \cr & \Rightarrow \tan \,\theta = \frac{{1 - \sqrt 7 }}{{1 + \sqrt 7 }} \cr} $$