Calculation:

As the given in the figure the railroad track and passes through the points \(\displaystyle{\left({11},\ {4}\right)}\) and \(\displaystyle{\left({8},\ {0}\right)}.\) Therefore slope is \(\displaystyle{\frac{{{0}-{4}}}{{{8}-{11}}}}={\frac{{-{4}}}{{-{3}}}}={\frac{{{4}}}{{{3}}}}.\)

As the new bike path is parallel to rail road track so slope are equal. Slope of new bike path is \(\displaystyle{\frac{{{4}}}{{{3}}}}.\)

Equation of new bike path is:

\(\displaystyle{y}-{5}={\frac{{{4}}}{{{3}}}}{\left({x}-{4}\right)}\)

\(\displaystyle{3}{y}-{15}={4}{x}-{16}\)

\(\displaystyle{4}{x}-{3}{y}={16}-{15}\)

\(\displaystyle{4}{x}-{3}{y}={1}\)

Thus, equation of the bike path is \(\displaystyle{4}{x}\ -\ {3}{y}={1}.\)

As the given in the figure the railroad track and passes through the points \(\displaystyle{\left({11},\ {4}\right)}\) and \(\displaystyle{\left({8},\ {0}\right)}.\) Therefore slope is \(\displaystyle{\frac{{{0}-{4}}}{{{8}-{11}}}}={\frac{{-{4}}}{{-{3}}}}={\frac{{{4}}}{{{3}}}}.\)

As the new bike path is parallel to rail road track so slope are equal. Slope of new bike path is \(\displaystyle{\frac{{{4}}}{{{3}}}}.\)

Equation of new bike path is:

\(\displaystyle{y}-{5}={\frac{{{4}}}{{{3}}}}{\left({x}-{4}\right)}\)

\(\displaystyle{3}{y}-{15}={4}{x}-{16}\)

\(\displaystyle{4}{x}-{3}{y}={16}-{15}\)

\(\displaystyle{4}{x}-{3}{y}={1}\)

Thus, equation of the bike path is \(\displaystyle{4}{x}\ -\ {3}{y}={1}.\)