Prove that the points A(7, 10), B(−2, 5) and C(3, −4)

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

$A B=\sqrt{(-2-7)^{2}+(5-10)^{2}}=\sqrt{(-9)^{2}+(-5)^{2}}=\sqrt{81+25}=\sqrt{106}$

$B C=\sqrt{(3-(-2))^{2}+(-4-5)^{2}}=\sqrt{(5)^{2}+(-9)^{2}}=\sqrt{25+81}=\sqrt{106}$

$A C=\sqrt{(3-7)^{2}+(-4-10)^{2}}=\sqrt{(-4)^{2}+(-14)^{2}}=\sqrt{16+196}=\sqrt{212}$

Since, AB and BC are equal, they form the vertices of an isosceles triangle.

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

and $(A C)^{2}=(\sqrt{212})^{2}=212$

Thus, (AB)2+(BC)2 = (AC)2

This show that $\triangle A B C$ is right- angled at $B$

Therefore, the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right-angled triangle.