What is the distance between $(- 3, \frac{5 π}{6})$ $(-3 ,(5 pi)/6 )$ and $(- 1, \frac{11 π}{12})$ $(-1 , (11 pi )/12 )$ ?

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What is the distance between $(- 3, \frac{5 π}{6})$ $(-3 ,(5 pi)/6 )$ and $(- 1, \frac{11 π}{12})$ $(-1 , (11 pi )/12 )$ ?

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Answer 1 · 2016-11-30T22:59:15

$d \approx 4.20$ $d~~4.20$

Explanation:

Each point on a polar plane is represented by the ordered pair $(r, θ)$ $(r,theta)$ .

So lets call the coordinates of $P_{1}$ $P_1$ as $(r_{1}, θ_{1})$ $(r_1,theta_1)$ and coordinates of $P_{2}$ $P_2$ as $(r_{2}, θ_{2})$ $(r_2,theta_2)$ . To find the distance between two points on a polar plane use the formula $d = \sqrt{{(r_{1})}_{1}^{2} + {(r_{2})}_{2}^{2} - 2 r_{1} r_{2} cos (θ_{2} - θ_{1})}$ $d=sqrt((r_1) ^2+(r_2)^2-2r_1r_2cos(theta_2-theta_1))$

Thererfore using the points $(- 3, \frac{5 π}{6})$ $(-3,(5pi)/6)$ and $(- 1, \frac{11 π}{12})$ $(-1,(11pi)/12)$ , and the formula

$d = \sqrt{{(r_{1})}_{1}^{2} + {(r_{2})}_{2}^{2} - 2 r_{1} r_{2} cos (θ_{2} - θ_{1})}$ $d=sqrt((r_1) ^2+(r_2)^2-2r_1r_2cos(theta_2-theta_1))$

we have

$d = \sqrt{{(- 3)}^{2} + {(- 1)}^{2} - 2 \cdot - 3 \cdot - 1 cos (\frac{11 π}{12} - \frac{5 π}{6})}$ $d=sqrt((-3)^2+(-1)^2-2*-3*-1cos((11pi)/12-(5pi)/6))$

$:. d~~4.20$

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What is the distance between $(- 3, \frac{5 π}{6})$ $(-3 ,(5 pi)/6 )$ and $(- 1, \frac{11 π}{12})$ $(-1 , (11 pi )/12 )$ ?

1 Answer

Explanation:

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