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

\(\displaystyle{D}=\sqrt{{{\left({y}_{{2}}-{y}_{{1}}\right)}^{{2}}+{\left({x}_{{2}}-{x}_{{1}}\right)}^{{2}}}}\)

\(\displaystyle=\sqrt{{{\left(-{\frac{{{5}}}{{\sqrt{{{2}}}}}}-{0}\right)}^{{2}}+{\left({\frac{{{5}}}{{\sqrt{{{2}}}}}}+{6}\right)}^{{2}}}}\)

\(\displaystyle=\sqrt{{{\frac{{{25}}}{{{2}}}}+{\frac{{{25}}}{{{2}}}}+{36}+{30}\sqrt{{{2}}}}}\)

\(\displaystyle=\sqrt{{{61}+{30}\sqrt{{{2}}}}}\)