Question

hallar la ecuacion de la recta en la forma normal hallar p y w x-3y+6=0

Answer 1

x - 3y + 6 = 0

A = 1

B= -3

C = 6

Ecuación normal = (Cos∝)x + (Sen∝)y - d = 0

$Coseno = \frac{A}{\sqrt{A^{2} } +B^{2} }\\Seno= \frac{B}{\sqrt{A^{2} } +B^{2} } \\\\d = \frac{C}{\sqrt{A^{2} } +B^{2} }$

$\sqrt{A^{2}+B^{2}}\\\\\sqrt{(1^{2}+(-3)^{2})}\\\\\sqrt{1+9} \\\\\sqrt{10}= 3,16$

$\frac{1}{3.16} x + (-\frac{3}{3.16} )y - (+\frac{6}{3.16} )= 0\\\\0,31x - 0.94y - 1.89 = 0$

Cos-1 (0.31) = ∝  |       Sen-1 (-0.94) = ∝

71.94° = ∝            |        -70.05° = ∝

(Cos71.94°)x  - (Sen-70.05°)y - 1.89 = 0