Find the rectangular coordinates of the pair of points (2,2π3) and (4,π6). Then find the distance, in simplified radical form, between the points.

The given polar coordinates are (r,θ)=(2,2π3) Since, the polar coordinates are x1=rcos⁡θ⇒2cos⁡(2π3) y1=rsin⁡θ⇒2sin⁡(2π3) x1=2(−0.5)=−1 y1=2(0.867)=1.734 Now, for another coordinate (4,π6) (r,θ)=(4,π6) Since, the polar coordinates are x2=rcos⁡θ4cos⁡(π6) y2=rsin⁡θ⇒4sin⁡(π6) x2=4(32)=23 y2=4(12)=2 So, the distance between the points is d=(x2−x1)2+(y2−y1)2 =(23+1)2+(2−1.734)2 =12+1+43+0.070756 =13.070756+43 Answer: Hence, the distance between the given points are =13.070756+43

Question: "Find the rectangular coordinates of the pair of..."

High School
Algebra
Algebra I
Exponents and radicals

DofotheroU

Answered question

2021-09-19

Find the rectangular coordinates of the pair of points

(2, 2 \frac{π}{3})

and

(4, \frac{π}{6})

. Then find the distance, in simplified radical form, between the points.

Answer & Explanation

Clelioo

Skilled2021-09-20Added 88 answers

The given polar coordinates are

(r, θ) = (2, \frac{2 π}{3})

Since, the polar coordinates are

x_{1} = r \cos θ \Rightarrow 2 \cos (\frac{2 π}{3})

y_{1} = r \sin θ \Rightarrow 2 \sin (\frac{2 π}{3})

x_{1} = 2 (- 0.5) = - 1

y_{1} = 2 (0.867) = 1.734

Now, for another coordinate

(4, \frac{π}{6})

(r, θ) = (4, \frac{π}{6})

Since, the polar coordinates are

x_{2} = r \cos θ 4 \cos (\frac{π}{6})

y_{2} = r \sin θ \Rightarrow 4 \sin (\frac{π}{6})

x_{2} = 4 (\frac{\sqrt{3}}{2}) = 2 \sqrt{3}

y_{2} = 4 (\frac{1}{2}) = 2

So, the distance between the points is

d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}

= \sqrt{{(2 \sqrt{3} + 1)}^{2} + {(2 - 1.734)}^{2}}

= \sqrt{12 + 1 + 4 \sqrt{3} + 0.070756}

= \sqrt{13.070756 + 4 \sqrt{3}}

Answer: Hence, the distance between the given points are

= \sqrt{13.070756 + 4 \sqrt{3}}

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