Write the equation √3x +y-4=0 in normal form to find the angle, Ø , that the normal to the line makes with the positive x-axis

Question

A. 45°

 

B. 90°

 

C. 60°

 

D. 30°

Answer 1

sqrt(3)*x + y - 4 = 0

 

sqrt(3)*dx + dy = 0 <--- implicit differentiation

 

sqrt(3) * dx = -dy

 

y' = dy/dx = - sqrt(3)

 

OR

 

y = -sqrt(3)*x + 4 <--- the  first derivative is -sqrt(3)

 

So the slope of the tangent line is -sqrt(3)

 

THe normal line, after rationalizing the denominator, is sqrt(3)/3 <--- the negative reciprocal

 

 

The original line crosses the x-axis at -4/-sqrt(3) = 4*Sqrt(3)/3 

 

So the normal line passes through ( 4*Sqrt(3)/3, 0) and has slope M=sqrt(3)/3

 

B = y - mx =  -   (sqrt(3)/3)*4*sqrt(3)/3 = 4/3

 

the equation of the normal line is y = sqrt(3)/3*X - 4/3

 

 

At x=3, the normal line is at 0.398717

 

the angle is inverse-tangent ( 0.39717 / ( 3 - (4/3)(sqrt(3)))

 

which is approximately 30 degrees