What is the distance between $(- 9, \frac{17}{12})$ $(-9 ,( 17 )/12 )$ and $(- 2, \frac{5 π}{4})$ $(-2 ,( 5 pi )/4 )$ ?

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What is the distance between $(- 9, \frac{17}{12})$ $(-9 ,( 17 )/12 )$ and $(- 2, \frac{5 π}{4})$ $(-2 ,( 5 pi )/4 )$ ?

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Answer 1 · 2016-11-30T22:55:20

see below

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 $(- 9, \frac{17}{12})$ $(-9,17/12)$ and $(- 2, \frac{5 π}{4})$ $(-2,(5pi)/4)$ , 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

$d = \sqrt{{(- 9)}^{2} + {(- 2)}^{2} - 2 \cdot - 9 \cdot - 2 cos (\frac{5 π}{4} - \frac{17}{12})}$ $d=sqrt((-9) ^2+(-2)^2-2*-9*-2cos((5pi)/4-17/12))$

$:. d~~10.68$

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What is the distance between $(- 9, \frac{17}{12})$ $(-9 ,( 17 )/12 )$ and $(- 2, \frac{5 π}{4})$ $(-2 ,( 5 pi )/4 )$ ?

1 Answer

Explanation:

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