What is the distance between $(2, \frac{7 π}{6})$ $(2 ,(7 pi)/6 )$ and $(- 7, \frac{5 π}{4})$ $(-7 , (5 pi )/4 )$ ?

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What is the distance between $(2, \frac{7 π}{6})$ $(2 ,(7 pi)/6 )$ and $(- 7, \frac{5 π}{4})$ $(-7 , (5 pi )/4 )$ ?

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Answer 1 · 2016-02-07T15:20:45

$4 \sqrt{5}$ $4sqrt5$

Explanation:

use the polar coordinate version of the distance formula

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

here let $(r_{1}, θ_{1}) = (2, \frac{7 π}{6}), (r_{2}, θ_{2}) = (- 7, \frac{5 π}{4})$ $(r_1,theta_1) = (2 ,(7pi)/6) , (r_2,theta_2) = (-7,(5pi)/4)$

substitute into formula

$d^{2} = 2^{2} + {(- 7)}^{2} - [2 \times 2 \times (- 7) \times cos (\frac{5 π}{4} - \frac{7 π}{6})]$ $d^2 = 2^2+(-7)^2 - [ 2 xx 2 xx(-7) xx cos((5pi)/4 - (7pi)/6)]$

$= 4 + 49 - [(- 28 \times cos (\frac{π}{12})]$ $= 4 + 49 -[(-28 xx cos(pi/12) ]$

= 53 - (-27) = 80

$d^{2} = 80 \Rightarrow d = \sqrt{80} = 4 \sqrt{5}$ $d^2 = 80 rArr d = sqrt80 =4sqrt5$

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What is the distance between $(2, \frac{7 π}{6})$ $(2 ,(7 pi)/6 )$ and $(- 7, \frac{5 π}{4})$ $(-7 , (5 pi )/4 )$ ?

1 Answer

Explanation:

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What is the distance between $(2, \frac{7 π}{6})$ $(2 ,(7 pi)/6 )$ and $(- 7, \frac{5 π}{4})$ $(-7 , (5 pi )/4 )$ ?