Question

reduce the equation x+y-root 2 =0 to the normal form x cos alpha +y sin alpha = p and hence find values of alpha and p

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{is}, \\ \mathrm{x}+\mathrm{y}-\sqrt{2}=0 \\ \Rightarrow \mathrm{x}+\mathrm{y}=\sqrt{2} \\ \Rightarrow \frac{\mathrm{x}}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2}}+\frac{\mathrm{y}}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2}}=\frac{\sqrt{2}}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2}} \\ \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{\mathrm{y}}{\sqrt{2}}=1 \\ \mathrm{So},\mathrm{we}\mathrm{get}\cos \mathrm{\alpha }=\frac{1}{\sqrt{2}}\mathrm{and}\sin \mathrm{\alpha }=\frac{1}{\sqrt{2}}\mathrm{and}\mathrm{p}=1 \\ \mathrm{Since},\cos \mathrm{\alpha }>0\mathrm{and}\sin \mathrm{\alpha }>\hspace{0.17em}0,\mathrm{so}\mathrm{\alpha }\mathrm{lies}\mathrm{in}\mathrm{first}\mathrm{quadrant}. \\ \mathrm{Now},\frac{\sin \mathrm{\alpha }}{\cos \mathrm{\alpha }}=\frac{1/\sqrt{2}}{1/\sqrt{2}} \\ \Rightarrow \tan \mathrm{\alpha }=1 \\ \Rightarrow \mathrm{\alpha }=45° \\ \mathrm{So},\mathrm{given}\mathrm{equation}\mathrm{in}\mathrm{normal}\mathrm{form}\mathrm{is}, \\ \boxed{\mathbf{x}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{45}\mathbf{°}\mathbf{}\mathbf{+}\mathbf{}\mathbf{y}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{45}\mathbf{°}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}} \end{gathered}$$