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Question

# The equations of one of the straight lines which pass through the point $\left(1,3\right)$ and make an angle ${\mathrm{tan}}^{-1}\sqrt{2}$ with the straight line, $y+1=3\sqrt{2}x$ is:

A

$5\sqrt{2}x+4y-15+4\sqrt{2}=0$

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B

$4\sqrt{2}x-5y-5+4\sqrt{2}=0$

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C

$4\sqrt{2}x+5y-4\sqrt{2}=0$

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D

$4\sqrt{2}x+5y-\left(15+4\sqrt{2}\right)=0$

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Solution

## The correct option is D $4\sqrt{2}x+5y-\left(15+4\sqrt{2}\right)=0$Explanation for correct option:Let the equation of the straight line is $y=mx+c$The straight line passes through $\left(1,3\right)$$\therefore 3=m+c$Given that the equations of one of the straight lines which pass through the point $\left(1,3\right)$ and make an angle ${\mathrm{tan}}^{-1}\sqrt{2}$ with the straight line, $y+1=3\sqrt{2}x$From the given figure,$\therefore \mathrm{tan}\left({\mathrm{tan}}^{-1}\sqrt{2}\right)=\left|\frac{\left(m-3\sqrt{2}\right)}{\left(1+3m\sqrt{2}\right)}\right|\phantom{\rule{0ex}{0ex}}⇒\sqrt{2}=\left|\frac{\left(m-3\sqrt{2}\right)}{\left(1+3m\sqrt{2}\right)}\right|$For positive $⇒\sqrt{2}=\frac{\left(m-3\sqrt{2}\right)}{\left(1+3m\sqrt{2}\right)}\phantom{\rule{0ex}{0ex}}⇒6m+\sqrt{2}=m-3\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒5m=-4\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒m=\frac{-4\sqrt{2}}{5}$For negative$⇒6m+\sqrt{2}=-m+3\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒7m=2\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒m=\frac{2\sqrt{2}}{7}$For $m=-\frac{4\sqrt{2}}{5}$$\therefore c=3-m\phantom{\rule{0ex}{0ex}}=\frac{15+4\sqrt{2}}{5}$$\therefore y=-\frac{4\sqrt{2}x}{5}+\frac{\left(15+4\sqrt{2}\right)}{5}\phantom{\rule{0ex}{0ex}}⇒4\sqrt{2}x+5y-\left(15+4\sqrt{2}\right)=0$For $m=\frac{2\sqrt{2}}{7}$$\therefore c=3-m\phantom{\rule{0ex}{0ex}}=\frac{21-2\sqrt{2}}{5}$$\therefore y=\frac{2\sqrt{2}x}{7}+\frac{\left(21-2\sqrt{2}\right)}{7}\phantom{\rule{0ex}{0ex}}⇒7y-2\sqrt{2}x-\left(21-2\sqrt{2}\right)=0$Hence, option $\left(\mathrm{D}\right)$ is correct.

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