What is the distance between $(5, \frac{7 π}{6})$ $(5 ,(7 pi)/6 )$ and $(3, \frac{- 7 π}{8})$ $(3 , (- 7 pi )/8 )$ ?

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Answer 1 · 2017-08-13T12:48:45

The distance between two points is $2.06$ $2.06$ unit.

First point conversion from polar coodinate to cartesian coordinate.

$r = 5, θ = \frac{7 \cdot π}{6} = 3.6652$ $r=5 , theta = (7*pi)/6=3.6652$ ,

x -cordinate is $r \cdot cos θ = 5 \cdot cos (3.6652) \approx - 4.33$ $r*cos theta =5 *cos(3.6652) ~~ -4.33$

y -cordinate is $r \cdot sin θ = 5 \cdot sin (3.6652) \approx - 2.5$ $r*sin theta =5 *sin(3.6652) ~~ -2.5$

Second point from polar coodinate to cartesian coordinate.

$r = 3, θ = 2 π - \frac{- 7 \cdot π}{8} = 3.5343$ $r=3 , theta = 2pi- (-7*pi)/8=3.5343$ ,

x -cordinate is $r \cdot cos θ = 3 \cdot cos (3.5343) \approx - 2.77$ $r*cos theta =3 *cos(3.5343) ~~ -2.77$

y -cordinate is $r \cdot sin θ = 3 \cdot sin (3.5343) \approx - 1.15$ $r*sin theta =3 *sin(3.5343) ~~ -1.15$

Cartesian cordinates of two points are $(- 4.33, - 2.5)$ $(-4.33, -2.5)$ and

# ( -2.77 , -1.15) . The distance between them is

$D = \sqrt{{(x_{1} - x_{2})}_{1}^{2} + {(y_{1} - y_{2})}_{1}^{2}}$ $D= sqrt ((x_1-x_2)^2+ (y_1-y_2)^2)$ or

$D = \sqrt{(- 4.33 - {(- 2.77)}^{2} + (- 2.5 - {(- 1.15)}^{2})}$ $D= sqrt ((-4.33-(-2.77)^2+ (-2.5-(-1.15)^2)$ or

$D \approx 2.06$ $D ~~ 2.06$ unit [Ans]

What is the distance between (5,7π6)(5 ,(7 pi)/6 ) and (3,−7π8)(3 , (- 7 pi )/8 )?