Question

If we reduce $3x+3y+7=0$ to the form $x\cos \alpha +y\sin \alpha =p$ then the value of p is
\[\begin{align}
  & A.\dfrac{7}{2\sqrt{3}} \\
 & B.\dfrac{7}{3} \\
 & C.\dfrac{3\sqrt{7}}{2} \\
 & D.\dfrac{7}{3\sqrt{2}} \\
\end{align}\]

Answer 1

Hint:At first, take 7 from left to right hand side, then multiply it by -1 as p should be positive. Then, divide it by $\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -3 \right)}^{2}}}\Rightarrow \sqrt{9+9}\Rightarrow \sqrt{18}\Rightarrow 3\sqrt{2}$ to get the required equation as value of $\cos \alpha \text{ and }\sin \alpha $ should be always between -1 and 1 to get the answer.

Complete step by step answer:
In the question, we are given an equation $3x+3y+7=0$ and we have to write it in form of $x\cos \alpha +y\sin \alpha =p$ and thus, find the value of p.
So, we are given the equation $3x+3y+7=0$ which can be further written as,
\[3x+3y=-7\]
The equation of form $x\cos \alpha +y\sin \alpha =p$ here p is the distance from the origin which cannot be considered as negative, so we have to multiply whole given re-written equation $3x+3y=-7$ by -1 so we get:
\[-3x-3y=7\]
We are not done as 0$\cos \alpha \text{ and }\sin \alpha $ cannot be -3 as their range lies between -1 and 1.
So, now we will divide the whole equation by $\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -3 \right)}^{2}}}\Rightarrow \sqrt{9+9}\Rightarrow \sqrt{18}\Rightarrow 3\sqrt{2}$ so it can be written as,
\[\dfrac{-3x-3y}{3\sqrt{2}}=\dfrac{7}{3\sqrt{2}}\]
Which we can write as,
\[\dfrac{-3x}{3\sqrt{2}}-\dfrac{3y}{3\sqrt{2}}=\dfrac{7}{3\sqrt{2}}\]
On further simplification we can write it as,
\[\left( \dfrac{-3}{3\sqrt{2}} \right)x+\left( \dfrac{-3}{3\sqrt{2}} \right)y=\dfrac{7}{3\sqrt{2}}\]
Now, cancelling term 3 on numerator and denominator, we have
\[\left( \dfrac{-1}{\sqrt{2}} \right)x+\left( \dfrac{-1}{\sqrt{2}} \right)y=\dfrac{7}{3\sqrt{2}}\]
So after writing the given equation in form of $x\cos \alpha +y\sin \alpha =p$ we can say that value of p on comparing equation is $\dfrac{7}{3\sqrt{2}}$
Thus, the correct option is D.

Note: Generally, while transforming equation in form of equation $x\cos \alpha +y\sin \alpha =p$ students tend to take p as a negative value, as they don't know that p is the distance of (x, y) from the origin. So, they should be careful about it.