How do you write the polar form of the equation of the line that passes through the points (4,-1) and (-2,3)?

Question

3084 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-30T17:09:40

Please see the explanation.

The slope, m, of the line is:

$m = \frac{3 - (- 1)}{- 2 - 4}$ $m = (3- (-1))/(-2-4)$

$m = \frac{4}{- 6}$ $m = 4/-6$

$m = - \frac{2}{3}$ $m = -2/3$

Use the point-slope form of the equation of a line:

$y = m (x - x_{1}) + y_{1}$ $y = m(x-x_1)+y_1$

$y = - \frac{2}{3} (x - 4) - 1$ $y = -2/3(x-4)-1$

Here is a graph of that line:

![Desmos.com]( useruploads.socratic.org )

Multiply both sides by 3:

$3 y = - 2 (x - 4) - 1$ $3y = -2(x-4)-1$

Distribute the -2:

$3 y = - 2 x + 8 - 1$ $3y = -2x+8-1$

Add 2x to both sides:

$2 x + 3 y = 7$ $2x+3y = 7$

Substitute $r cos (θ)$ $rcos(theta)$ for x and $r sin (θ)$ $rsin(theta)$ for y:

$2 r cos (θ) + 3 r sin (θ) = 7$ $2rcos(theta)+3rsin(theta) = 7$

Factor out r:

$r (2 cos (θ) + 3 sin (θ)) = 7$ $r(2cos(theta)+3sin(theta)) = 7$

Divide both sides by $(2 cos (θ) + 3 sin (θ))$ $(2cos(theta)+3sin(theta))$

$r = \frac{7}{2 cos (θ) + 3 sin (θ)}$ $r = 7/(2cos(theta)+3sin(theta))$

Here is a graph of that equation:

![Desmos.com]( useruploads.socratic.org )