$\text { Find a polar equation in the form } r \cos \left( \theta - \theta _ { 0 } \right) = r _ { 0 } \text { for the given line. }$
-$\sqrt { 3 } x + y = 4$
A) $\mathrm { r } \cos \left( \theta - \frac { \pi } { 3 } \right) = 2$
B) $r \cos \left( \theta + \frac { 2 \pi } { 3 } \right) = 2$
C) $r \cos \left( \theta + \frac { \pi } { 6 } \right) = 2$
D) $r \cos \left( \theta - \frac { \pi } { 6 } \right) = 2$

Question

Quizplus · Accepted Answer

We can rewrite the equation as $y = 4-\sqrt{3}x$, so the $y$-intercept is 4 and the slope is $-\sqrt{3}$. Therefore, the angle between the line and the $x$-axis is $\frac{\pi}{6}$ clockwise. We want a polar equation of the form $r\cos(	heta-	heta_0) = r_0$, where $	heta_0$ is the angle from the polar axis to the line. Since the line makes an angle of $\frac{\pi}{6}$ clockwise from the polar axis, we have $	heta_0 = \frac{\pi}{6}+\pi = \frac{7\pi}{6}$ (adding $\pi$ to account for the clockwise rotation). We also know that $r_0$ is the distance from the origin to the $y$-intercept, which is 4. Therefore, our polar equation is $r\cos\left(	heta-\frac{7\pi}{6}ight) = 4$. We can simplify this to $r\cos\left(	heta-\frac{\pi}{6}ight) = 2$, so the answer is $\boxed{	ext{(D)}}$.

Find a polar equation in the form rcos⁡(θ−θ0)=r0 for the given line. \text { Find a polar equation in the form } r \cos \left( \theta - \theta _ { 0 } \right) = r _ { 0 } \text { for the given line. } Find a polar equation in the form rcos(θ−θ0​)=r0​ for the given line.

$\text { Find a polar equation in the form } r \cos \left( \theta - \theta _ { 0 } \right) = r _ { 0 } \text { for the given line. }$