Slope of a line passing through

Question:

Slope of a line passing through $\mathrm{P}(2,3)$ and intersecting the line $x+y=7$ at a distance of 4 units from $P$, is:

  1. (1) $\frac{1-\sqrt{5}}{1+\sqrt{5}}$

  2. (2) $\frac{1-\sqrt{7}}{1+\sqrt{7}}$

  3. (3) $\frac{\sqrt{7}-1}{\sqrt{7}+1}$

  4. (4) $\frac{\sqrt{5}-1}{\sqrt{5}+1}$


Correct Option: , 2

Solution:

Since point at 4 units from $\mathrm{P}(2,3)$ will be

$\mathrm{A}(4 \cos \theta+2,4 \sin (\theta+3)$ and this point will satisfy

the equation of line $x+y=7$

$\Rightarrow \cos \theta+\sin \theta=\frac{1}{2}$

On squaring

$\Rightarrow \sin 2 \theta-\frac{3}{4} \Rightarrow \frac{2 \tan \theta}{1+\tan ^{2} \theta}=-\frac{3}{4}$

$\Rightarrow 3 \tan ^{2} \theta+8 \tan \theta+3=0$

$\Rightarrow \tan \theta=\frac{-8 \pm 2 \sqrt{7}}{6} \quad($ ignoring $-$ ve sign $)$

$\Rightarrow \tan \theta=\frac{-8+2 \sqrt{7}}{6}=\frac{1-\sqrt{7}}{1+\sqrt{7}}$

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