Find the rectangular coordinates of the pair of points (6,π) and (5,7π4). Then find the distance, in simplified radical form, between the points.

The given points are (6,π) and (5,7π4) Known fact: x=rcos⁡θ y=rsin⁡θ Calculation: For (6,π) x=6cos⁡π =6(−1) =−6 y=6sin⁡π =6(0) =0 Thus (x,y)=(−6,0) For (5,7π4) x=5cos⁡7π4 =5(12) =52 y=5sin⁡7π4 =−52 Thus, (x,y)=(52,−52) Now find the distance D=(y2−y1)2+(x2−x1)2 =(−52−0)2+(52+6)2 =252+252+36+302 =61+302

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High School
Algebra
Algebra I
Exponents and radicals

Nann

Answered question

2021-10-01

Find the rectangular coordinates of the pair of points

(6, π)

and

(5, 7 \frac{π}{4})

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

Answer & Explanation

Faiza Fuller

Skilled2021-10-02Added 108 answers

The given points are
$(6, π)$ and $(5, \frac{7 π}{4})$
Known fact:
$x = r \cos θ$
$y = r \sin θ$
Calculation:
For $(6, π)$
$x = 6 \cos π$
$= 6 (- 1)$
$= - 6$
$y = 6 \sin π$
$= 6 (0)$
$= 0$
Thus $(x, y) = (- 6, 0)$
For $(5, \frac{7 π}{4})$
$x = 5 \cos \frac{7 π}{4}$
$= 5 (\frac{1}{\sqrt{2}})$
$= \frac{5}{\sqrt{2}}$
$y = 5 \sin \frac{7 π}{4}$
$= - \frac{5}{\sqrt{2}}$
Thus, $(x, y) = (\frac{5}{\sqrt{2}}, - \frac{5}{\sqrt{2}})$
Now find the distance
$D = \sqrt{{(y_{2} - y_{1})}^{2} + {(x_{2} - x_{1})}^{2}}$
$= \sqrt{{(- \frac{5}{\sqrt{2}} - 0)}^{2} + {(\frac{5}{\sqrt{2}} + 6)}^{2}}$
$= \sqrt{\frac{25}{2} + \frac{25}{2} + 36 + 30 \sqrt{2}}$
$= \sqrt{61 + 30 \sqrt{2}}$

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