An isosceles triangle has at least two sides of equal length. Let's determine the length of each side using the distance formula:

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

Substitute for AB:

\(\displaystyle{A}{B}=\sqrt{{{\left({3}-{0}\right)}^{{2}}+{\left({4}-{0}\right)}^{{2}}}}=\sqrt{{{9}+{16}}}=\sqrt{{25}}={5}\)

Substitute for BC:

\(\displaystyle{B}{C}=\sqrt{{{\left({7}-{3}\right)}^{{2}}+{\left({1}-{4}\right)}^{{2}}}}=\sqrt{{{16}+{9}}}=\sqrt{{25}}={5}\)

Since \(\displaystyle{A}{B}={B}{C},\triangle{A}{B}{C}\) is isosceles.

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

Substitute for AB:

\(\displaystyle{A}{B}=\sqrt{{{\left({3}-{0}\right)}^{{2}}+{\left({4}-{0}\right)}^{{2}}}}=\sqrt{{{9}+{16}}}=\sqrt{{25}}={5}\)

Substitute for BC:

\(\displaystyle{B}{C}=\sqrt{{{\left({7}-{3}\right)}^{{2}}+{\left({1}-{4}\right)}^{{2}}}}=\sqrt{{{16}+{9}}}=\sqrt{{25}}={5}\)

Since \(\displaystyle{A}{B}={B}{C},\triangle{A}{B}{C}\) is isosceles.