JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 5)
The equation of one of the straight lines which passes through the point (1, 3) and makes an angles $${\tan ^{ - 1}}\left( {\sqrt 2 } \right)$$ with the straight line, y + 1 = 3$${\sqrt 2 }$$ x is :
$$4\sqrt 2 x + 5y - \left( {15 + 4\sqrt 2 } \right) = 0$$
$$5\sqrt 2 x + 4y - \left( {15 + 4\sqrt 2 } \right) = 0$$
$$4\sqrt 2 x + 5y - 4\sqrt 2 = 0$$
$$4\sqrt 2 x - 5y - \left( {5 + 4\sqrt 2 } \right) = 0$$
Explanation
Let slope of line be m
$$ \therefore $$ $$\left| {{{m - 3\sqrt 2 } \over {1 + 3\sqrt 2 m}}} \right| = \sqrt 2 $$
$$ \Rightarrow m - 3\sqrt 2 = \pm \,\sqrt 2 \pm 6m$$
$$ \Rightarrow m \mp 6m = \pm \sqrt 2 + 3\sqrt 2 $$
$$ \Rightarrow m = - {{4\sqrt 2 } \over 5}$$ or $${{2\sqrt 2 } \over 7}$$
Hence line can be
$$y - 3 = {{ - 4\sqrt 2 } \over 5}(x - 1)$$
$$ \Rightarrow 5y - 15 = - 4\sqrt 2 x + 4\sqrt 2 $$
$$ \Rightarrow 4\sqrt 2 x + 5y - (15 + 4\sqrt 2 ) = 0$$
$$ \therefore $$ $$\left| {{{m - 3\sqrt 2 } \over {1 + 3\sqrt 2 m}}} \right| = \sqrt 2 $$
$$ \Rightarrow m - 3\sqrt 2 = \pm \,\sqrt 2 \pm 6m$$
$$ \Rightarrow m \mp 6m = \pm \sqrt 2 + 3\sqrt 2 $$
$$ \Rightarrow m = - {{4\sqrt 2 } \over 5}$$ or $${{2\sqrt 2 } \over 7}$$
Hence line can be
$$y - 3 = {{ - 4\sqrt 2 } \over 5}(x - 1)$$
$$ \Rightarrow 5y - 15 = - 4\sqrt 2 x + 4\sqrt 2 $$
$$ \Rightarrow 4\sqrt 2 x + 5y - (15 + 4\sqrt 2 ) = 0$$
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