What is the distance between $(- 6, \frac{π}{3})$ $(-6 , pi/3 )$ and $(7, \frac{π}{4})$ $(7 , pi/4 )$ ?

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What is the distance between $(- 6, \frac{π}{3})$ $(-6 , pi/3 )$ and $(7, \frac{π}{4})$ $(7 , pi/4 )$ ?

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Answer 1 · 2016-12-01T19:49:29

$d \approx 12.89$ $d~~12.89$

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 $(- 6, \frac{π}{3})$ $(-6,(pi)/3)$ and $(7, \frac{π}{4})$ $(7,(pi)/4)$ , 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

$d = \sqrt{{(- 6)}^{2} + {(7)}^{2} - 2 \cdot - 6 \cdot 7 cos (\frac{π}{4} - \frac{π}{3})}$ $d=sqrt((-6)^2+(7)^2-2*-6*7cos((pi)/4-(pi)/3))$

$:. d~~12.89$

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What is the distance between $(- 6, \frac{π}{3})$ $(-6 , pi/3 )$ and $(7, \frac{π}{4})$ $(7 , pi/4 )$ ?

1 Answer

Explanation:

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What is the distance between (-6 , pi/3 )(−6,π3)(-6 , pi/3 ) and (7 , pi/4 )(7,π4)(7 , pi/4 )?

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What is the distance between $(- 6, \frac{π}{3})$ $(-6 , pi/3 )$ and $(7, \frac{π}{4})$ $(7 , pi/4 )$ ?