Polar equation of a line

TEMPORARY PAGE - I'm working it out now so may as well use the steps

Line

• [ilmath]y=mx+c[/ilmath]

Parametric

• [math]r=\sqrt{t^2(m^2+1)+2mtc+c^2}[/math]
• [math]\theta=\arctan\left(m+\frac{c}{t}\right)[/math]

Polar form

Method:

• Substitute:
  • [math]t=\frac{c}{\tan(\theta)-m}[/math] into [math]r=\sqrt{t^2(m^2+1)+2mtc+c^2}[/math]

Working:

• [math]r=\sqrt{\left(\frac{c}{\tan(\theta)-m}\right)^2(m^2+1)+2mc\left(\frac{c}{\tan(\theta)-m}\right)+c^2}[/math]
• [math]r=|c|\sqrt{\frac{m^2+1}{(\tan(\theta)-m)^2}+\frac{2m}{\tan(\theta)-m}+1}[/math]
  • Solving the Quadratic equation (treating [ilmath]\frac{1}{\tan(\theta)-m} [/ilmath] as the variable we are quadratic in we see)
    • [ilmath]a=m^2+1[/ilmath]
    • [ilmath]b=2m[/ilmath]
    • [ilmath]c=1[/ilmath]
  • To get roots [ilmath]\frac{\pm j-m}{m^2+1} [/ilmath] where [ilmath]j=\sqrt{-1}[/ilmath]
  • noting that [ilmath]m^2+1=(m-j)(m+j)[/ilmath] we see that the roots are simply [ilmath]\frac{-1}{m\pm j} [/ilmath]
  • this means [math](x-r_1)(x-r_2) = 0[/math] whenever [ilmath]x=r_1[/ilmath] or [ilmath]x=r_2[/ilmath], so:
• [math]r=|c|\sqrt{\left( \frac{1}{\tan(\theta)-m} + \frac{1}{m-j} \right)\left( \frac{1}{\tan(\theta)-m} + \frac{1}{m+j} \right)}[/math]

Questions

• Is this the best form?
• if [ilmath]m=1[/ilmath] and [ilmath]c=1[/ilmath] then
  • - this can't be the only special case!