Question

If we reduce $3x + 3y + 7 = 0$ in the form of $x\cos \alpha + y\sin \alpha = p$, then the value of P is?
$
  A)\dfrac{7}{{2\sqrt 3 }} \\
  B)\dfrac{7}{3} \\
  C)\dfrac{{3\sqrt 7 }}{2} \\
  D)\dfrac{7}{{3\sqrt 2 }} \\
$

Answer 1

Hint: Find the common angle in sine and cosine and then factorize the equation to get in the required form and compare to get the value.
First, we initially send out terms to the other side and then since the required form has P, it is the distance from the origin and it cannot be negative, we try to put in the positive value by multiplying minus on both sides. Then we find out the angle at which the value is there and then we obtain the required form and compare it, hence finding the value of P.

Complete step by step solution:
Given an equation
$3x + 3y + 7 = 0$
To get the required value, we send the constant to the right
$3x + 3y = - 7$
The RHS, cannot be negative, as it is the distance from the origin, so we multiply negative on both sides.
$
   - (3x + 3y) = - ( - 7) \\
   - 3x - 3y = 7 \\
$
Since, we don’t have a value of $\sin x$ and $\cos x$ which has -3, so we divide both sides $\sqrt {{a^2} + {b^2}} $.
\[
   = \sqrt {{{( - 3)}^2} + {{( - 3)}^2}} \\
   = \sqrt {18} \\
   = 3\sqrt 2 \\
\]
We are going to divide both side by \[ = 3\sqrt 2 \]
\[\dfrac{{ - 3x}}{{3\sqrt 2 }} - \dfrac{{3y}}{{3\sqrt 2 }} = \dfrac{7}{{3\sqrt 2 }}\]
Now, cancelling like terms and then we get,
\[\dfrac{{ - x}}{{\sqrt 2 }} - \dfrac{y}{{\sqrt 2 }} = \dfrac{7}{{3\sqrt 2 }}\]
If we compare with the given form, we get
$\sin \alpha = - \dfrac{1}{{\sqrt 2 }}$
$\cos \alpha = - \dfrac{1}{{\sqrt 2 }}$
Since values for these angles exist, we don’t need to simplify the equation, as we have converted the expression to the required form,
Which is
\[\dfrac{{ - x}}{{\sqrt 2 }} - \dfrac{y}{{\sqrt 2 }} = \dfrac{7}{{3\sqrt 2 }}\]
When we compare the above equation with $x\cos \alpha + y\sin \alpha = p$, We get the value of P, which is
$P = \dfrac{7}{{3\sqrt 2 }}$

So, the correct answer is Option D.

Note: For the given $x\cos \alpha + y\sin \alpha = p$, p should always be positive, as it is the distance from the origin. Also, since both $\operatorname{Sin} $ and $\operatorname{Cos} $ have the same angle $\operatorname{Cos} $. They should be factorized accordingly to get the same and equal angles.