Question

The perpendicular from of the line with slope $m=\frac{1}{\sqrt{2}}$ and distance between origin to the line is 4 then the equation is:

• A. $\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=4$
• B. x+y=16$\sqrt{2}$
• C. x-y=16$\sqrt{2}$
• D. None of these

Answer 1

The correct option is A $\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=4$

$m=\frac{1}{\sqrt{2}}=\tan \theta$

$\Rightarrow$ inclination ${}^{\prime }{\theta }^{\prime }={45}^o$ and perpendicular distance from origin to the line is $4$.

Using normal form equation of straight line is $x\cos \alpha +y\sin \alpha =P$

$\Rightarrow x\cos {45}^o+y\sin {45}^o=4$

$\Rightarrow \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=4$.