This is the problem and the answer given by my book:
My solution:
$3x+\sqrt{3}y+2=0…(i)$
$x\cos\alpha+y\sin\alpha=p…(ii)$
Since (i) & (ii) are equations of the same straight line,
$$\frac{3}{\cos\alpha}=\frac{\sqrt{3}}{\sin\alpha}=\frac{2}{-p}$$
$$\implies -3p=2\cos\alpha…(i)$$
$$\implies -\sqrt{3}p=2\sin\alpha…(ii)$$
$(i)^2+(ii)^2:-$
$$9p^2+3p^2=4\cos^2\alpha+4\sin^2\alpha$$
$$\implies 12p^2=4$$
$$\implies p^2=\frac{1}{3}$$
$$\implies \sqrt{p^2}=\sqrt{\frac{1}{3}}$$
$$\implies |p|=\frac{1}{\sqrt{3}}$$
$$\implies p=\pm \frac{1}{\sqrt{3}}$$
Why did the book pick only the negative value of $p$?
Best Answer
You are correct that $p$ needn't be negative: $$3x+\sqrt{3}y+2=0,$$ $$x\cos\left(\frac{\pi}{6}\right)+y\sin\left(\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}},$$ and $$x\cos\left(\frac{7\pi}{6}\right)+y\sin\left(\frac{7\pi}{6}\right)=\frac{1}{\sqrt{3}}$$ are all represented by the same line.