Question

If we reduce $3x+3y+7=0$ to the form $x\cos \alpha +y\sin \alpha =p$, then the value of $p$ is

• A. $\frac{7}{2\sqrt{3}}$
• B. $\frac{7}{3}$
• C. $\frac{7}{3\sqrt{2}}$
• D. $\frac{3\sqrt{7}}{2}$

Answer 1

The correct option is C

$\frac{7}{3\sqrt{2}}$

Explanation for the correct answer:

Finding the value of $p$:

Given,

$3x+3y+7=0$

this can be written as,

$-x-y=\frac{7}{3}$

Divide by $\sqrt{2}$on both sides, we get

$$\begin{gathered} \frac{-\mathrm{x}}{\sqrt{2}}+\frac{-\mathrm{y}}{\sqrt{2}}=\frac{7}{3\sqrt{2}} \\ \Rightarrow \mathrm{x}\cos \left(225\right)+\mathrm{y}\sin \left(225\right)=\frac{7}{3\sqrt{2}}\left[\mathbf{\because }\mathbf{cos}\mathbf{}\mathbf{225}\mathbf{=}\mathbf{sin}\mathbf{}\mathbf{225}\mathbf{=}\frac{\mathbf{-}\mathbf{1}}{\sqrt{\mathbf{2}}}\right] \end{gathered}$$

On comparing the above equation with $x\cos \alpha +y\sin \alpha =p$ ,we get the value of $p$ as $\frac{7}{3\sqrt{2}}$

Hence, the correct option is D.