Question

The Normal form of the line ${\text{x}}\,{\text{ + }}\,{\text{y}}\,{\text{ + }}\,\sqrt {\text{2}} {\text{ = 0}}$ is

Answer 1

Hint: Generally a normal form of the line is in the form.
${\text{x}}\cos \,\theta \, + \,{\text{y}}\,{\text{sin }}\theta \,{\text{ = }}\,{\text{1}}{\text{.}}$
So we need to express the given equation ${\text{x}}\,{\text{ + }}\,{\text{y}}\, + \,\sqrt 2 \, = \,0$ in the normal from

Complete step-by-step answer:
General form is given as:

{\text{x}}\cos \,\theta \, + \,{\text{y}}\,{\text{sin }}\theta \,{\text{ = }}\,{\text{1}} \\\ {\text{x}}\,{\text{ + }}\,{\text{y}}\,{\text{ + }}\,\sqrt 2 \, = \,0 \\\ {\text{step}} - 1 \\\ {\text{x}}\,{\text{ + }}\,{\text{y}}\, = \, - \sqrt 2 \\\ {\text{step}}\,{\text{ - 2, }} \\\ {\text{divide}}\,{\text{by}}\,' - \sqrt 2 ' \\\ (x)\left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)\, + \,({\text{y}})\left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)\, = \,1 \\\ \end{gathered} $$ Now compare with standard equation $$\cos \theta \, = \,\dfrac{-1}{{\sqrt 2 }};\,\sin \theta \, = \,\dfrac{{ - 1}}{{\sqrt 2 }}$$ Both $\cos \,\theta $ & $\sin \theta $ takes $'\dfrac{{ - 1}}{{\sqrt 2 }}'$ in 3rd Quadrant, $\therefore \,\theta \, = \,\dfrac{{5\pi }}{4}$ $\therefore \,{\text{x}}\,\cos \,\dfrac{{5\pi }}{4}\, + \,{\text{y}}\,\sin \dfrac{{5\pi }}{4}\, = \,1$ Note: The common mistake can be made here is taking the wrong value of ‘’. Like need to be careful of the sign values in respective quadrants.