Question

Transform the equation 3x + 2y + 12 = 0 into normal form

Answer 1

normal form
3x + 2y = -12
now ,

$\sqrt{(a) {}^{2} } + (b) {}^{2} \\ \sqrt{9 +4 } \\ \sqrt{13}$
now dividing the both sides by root 13 we get
$\frac{3x}{ \sqrt{13} } + \frac{2y}{\sqrt{13} } = \frac{ - 12}{ \sqrt{13} } \\ \\ x \times \frac{3}{ \sqrt{13} } + y \times \frac{3}{ \sqrt{13} } = \frac{ - 12}{\sqrt{13} } \\ \\ x \: cos \alpha + y \: sin \alpha = p$

where cos
$\cos\alpha = \frac{3}{ \sqrt{13} } \\ \sin( \alpha ) = \frac{2}{\sqrt{13} } \\ p = \frac{ - 12}{\sqrt{13} }$