Straight Line and Pair of Straight Lines 1 Question 6
6. Slope of a line passing through $P(2,3)$ and intersecting the line, $x+y=7$ at a distance of 4 units from $P$, is
(2019 Main, 9 April I)
(a) $\frac{1-\sqrt{5}}{1+\sqrt{5}}$
(b) $\frac{\sqrt{7}-1}{\sqrt{7}+1}$
(c) $\frac{1-\sqrt{7}}{1+\sqrt{7}}$
(d) $\frac{\sqrt{5}-1}{\sqrt{5}+1}$
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Solution:
- Let the slope of line is $m$, which is passing through $P(2,3)$.
Since, the distance of a point $\left(x _1, y _1\right)$ from the line $a x+b y+c=0$ is $d=\left|\frac{a x _1+b y _1+c}{\sqrt{a^{2}+b^{2}}}\right|$.
$\therefore$ The distance of a point $P(2,3)$ from the line $x+y-7=0$, is
$$ d=\frac{|2+3-7|}{\sqrt{1+1}}=\frac{2}{\sqrt{2}}=\sqrt{2} $$
Now, in $\triangle P R Q$,
$$ Q R=\sqrt{16-d^{2}}=\sqrt{16-2}=\sqrt{14} $$
$$ \begin{array}{llrl} & \therefore & \tan \theta=\frac{d}{Q R} & =\frac{\sqrt{2}}{\sqrt{14}}=\frac{1}{\sqrt{7}}=\left|\frac{m+1}{1-m}\right| \\ \Rightarrow & \frac{m+1}{1-m} & = \pm \frac{1}{\sqrt{7}} \\ \Rightarrow & \frac{m+1}{1-m} & =\frac{1}{\sqrt{7}} \text { or } \frac{m+1}{1-m}=-\frac{1}{\sqrt{7}} \\ \Rightarrow & m & =\frac{1-\sqrt{7}}{1+\sqrt{7}} \text { or } m=\frac{-1-\sqrt{7}}{\sqrt{7}-1} \end{array} $$
