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. $\dfrac{7}{2\sqrt{3}}$
B. $\dfrac{7}{3}$
C. $\dfrac{3\sqrt{7}}{2}$
D. $\dfrac{7}{3\sqrt{2}}$

Answer 1

In this problem, we need to find the value of “p” given that we reduce $3x+3y+7=0$ to the form $x\cos \alpha +y\sin \alpha =p$ . We first move $7$ to the RHS and then multiply the equation by $-1$ . Upon comparing, we find that the coefficients of x and y are same. So, sine and cosine must be same and equal to $-\dfrac{1}{\sqrt{2}}$ . This gives the value of p as $\dfrac{7}{3\sqrt{2}}$ .

Complete step-by-step answer:
We are given the linear equation $3x+3y+7=0$ . We are said that this equation can be expresses in some other form and that form is $x\cos \alpha +y\sin \alpha =p$ . The two equations thus are equivalent. But, if we compare the two given equations, we do not find anything common or comparable. The coefficient of “x” in the first equation is $3$ and that in the second equation is $\cos \alpha$ . Similarly, the coefficient of “y” in the first equation is $sin\alpha$ and that in the second equation is $3$ . This clearly implies that the two equations are not directly comparable. This is because the range of sine and cosine functions is $\left[ -1,1 \right]$ . But, the first equation says that it is $3$ . This is irrelevant. So, we have to think of something else.
Let us move $7$ to the RHS of the equation and multiply by $-1$ . We get,
$-3x-3y=7....\left( i \right)$
We notice that the coefficients of “x” and “y” in the first equation are same. This must also be the case in the second equation. So,
$\begin{aligned} & \Rightarrow \cos \alpha =\sin \alpha \\\ & \Rightarrow \alpha =-\dfrac{\pi }{4} \\\ & \Rightarrow \cos \alpha =\sin \alpha =-\dfrac{1}{\sqrt{2}} \\\ \end{aligned}$
We now divide the two sides of the equation (i) by $3\sqrt{2}$ . This gives,
$\begin{aligned} & \Rightarrow -\dfrac{3}{3\sqrt{2}}x-\dfrac{3}{3\sqrt{2}}y=\dfrac{7}{3\sqrt{2}} \\\ & \Rightarrow -\dfrac{1}{\sqrt{2}}x-\dfrac{1}{\sqrt{2}}y=\dfrac{7}{3\sqrt{2}} \\\ \end{aligned}$
Thus, we can conclude that the value of “p” will be $\dfrac{7}{3\sqrt{2}}$ which is option d.

So, the correct answer is “Option d”.

Note: The most important thing that we must take care of in this problem is the sign. We must multiply the two sides by $-1$ in order to match with the answers. Also, we must note the fact that we have drawn the inference that $\cos \alpha =\sin \alpha$ only because the coefficients are the same.