JEE Main 2019 (Online) 9th April Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 85 English Explanation

We know parametric form of straight line is

$${{x - {x_1}} \over {\cos \theta }} = {{y - {y_1}} \over {\sin \theta }} = r$$

$$ \Rightarrow $$ $${{x - 2} \over {\cos \theta }} = {{y - 3} \over {\sin \theta }} = 4$$

$$ \therefore $$ x = 4 + 2cos$$\theta $$

y = 4 + 3sin$$\theta $$

$$ \therefore $$ Point A = (4 + 2cos$$\theta $$, 4 + 3sin$$\theta $$)

Point A lies on line x + y = 7,

$$ \therefore $$ (4 + 2cos$$\theta $$) + (4 + 3sin$$\theta $$) = 7

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

$$ \Rightarrow $$ $${\sin ^2}\theta + {\cos ^2}\theta $$ + 2sin$$\theta $$cos$$\theta $$ = $${1 \over 4}$$

$$ \Rightarrow $$ 1 + sin 2$$\theta $$ = $${1 \over 4}$$

$$ \Rightarrow $$ sin 2$$\theta $$ = $$ - {3 \over 4}$$

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

$$ \Rightarrow $$ 3tan²$$\theta $$ + 8tan$$\theta $$ + 3 = 0

$$ \Rightarrow $$ tan$$\theta $$ = $${{ - 8 \pm 2\sqrt 7 } \over 6}$$

So slope = $${{ - 8 \pm 2\sqrt 7 } \over 6}$$

By checking each options,

Slope = $${{1 - \sqrt 7 } \over {1 + \sqrt 7 }}$$

As $${{1 - \sqrt 7 } \over {1 + \sqrt 7 }}$$ = $${{{{\left( {1 - \sqrt 7 } \right)}^2}} \over {1 - 7}}$$ = $${{8 - 2\sqrt 7 } \over { - 6}}$$ = $${{ - 8 + 2\sqrt 7 } \over 6}$$