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

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Answer 1 · 2017-05-30T17:09:40

Please see the explanation.

The slope, m, of the line is:

$m = (3- (-1))/(-2-4)$

$m = 4/-6$

$m = -2/3$

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

$y = m(x-x_1)+y_1$

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

Here is a graph of that line:

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

Multiply both sides by 3:

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

Distribute the -2:

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

Add 2x to both sides:

$2x+3y = 7$

Substitute $rcos(theta)$ for x and $rsin(theta)$ for y:

$2rcos(theta)+3rsin(theta) = 7$

Factor out r:

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

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

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

Here is a graph of that equation:

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