The given points are $A(7,10), B(-2,5)$ and $C(3,-4)$.

$$
\begin{aligned}
&\mathrm{AB}=\sqrt{(-2-7)^{2}+(5-10)^{2}}=\sqrt{(-9)^{2}+(-5)^{2}}=\sqrt{81+25}=\sqrt{106} \\
&\mathrm{BC}=\sqrt{(3-(-2))^{2}+(-4-5)^{2}}=\sqrt{(5)^{2}+(-9)^{2}}=\sqrt{25+81}=\sqrt{106} \\
&\mathrm{AC}=\sqrt{(3-7)^{2}+(-4-10)^{2}}=\sqrt{(-4)^{2}+(-14)^{2}}=\sqrt{16+196}=\sqrt{212}
\end{aligned}
$$

Since, $A B$ and $B C$ are equal, they form the vertices of an isosceles triangle

$(A B)^{2}+(B C)^{2}=(\sqrt{106})^{2}+(\sqrt{106})^{2}=212$

and $(\mathrm{AC})^{2}=(\sqrt{212})^{2}=212$.

Thus, $(A B)^{2}+(B C)^{2}=(A C)^{2}$

=> show that $\triangle \mathrm{ABC}$ is right- angled at $\mathrm{B}$.

Therefore, the points $\mathrm{A}(7,10), \mathrm{B}(-2,5)$ and $\mathrm{C}(3,-4)$ are the vertices of an isosceles rightangled triangle.