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

Question

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

1 Answer

Impact of this question

1648 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-01T17:39:53

$d \approx 6.76$ $d~~6.76$

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 $(2, \frac{7 π}{6})$ $(2,(7pi)/6)$ 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

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

$:. d~~6.76$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the distance between (2 ,(7 pi)/6 )(2,7π6)(2 ,(7 pi)/6 ) and (-7 , (- pi )/4 )(−7,−π4)(-7 , (- pi )/4 )?

1 Answer

Explanation:

Related questions

Impact of this question

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