Find the rectangular coordinates of the pair of points (6, \pi) and (5,

Nann 2021-10-01 Answered
Find the rectangular coordinates of the pair of points \(\displaystyle{\left({6},\pi\right)}\) and \(\displaystyle{\left({5},{7}\frac{\pi}{{4}}\right)}\). Then find the distance, in simplified radical form, between the points.

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Expert Answer

Faiza Fuller
Answered 2021-10-02 Author has 18501 answers

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

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