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

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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Faiza Fuller

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
$$\displaystyle{D}=\sqrt{{{\left({y}_{{2}}-{y}_{{1}}\right)}^{{2}}+{\left({x}_{{2}}-{x}_{{1}}\right)}^{{2}}}}$$
$$\displaystyle=\sqrt{{{\left(-{\frac{{{5}}}{{\sqrt{{{2}}}}}}-{0}\right)}^{{2}}+{\left({\frac{{{5}}}{{\sqrt{{{2}}}}}}+{6}\right)}^{{2}}}}$$
$$\displaystyle=\sqrt{{{\frac{{{25}}}{{{2}}}}+{\frac{{{25}}}{{{2}}}}+{36}+{30}\sqrt{{{2}}}}}$$
$$\displaystyle=\sqrt{{{61}+{30}\sqrt{{{2}}}}}$$