Prove that the points (7, 10),

Let the given points be A(7, 10), B(−2, 5) and C(3, −4).

Using distance formula, we have

$\mathrm{AB}=\sqrt{(-2-7)^{2}+(5-10)^{2}}=\sqrt{(-9)^{2}+(-5)^{2}}=\sqrt{81+25}=\sqrt{106}$ units

$\mathrm{BC}=\sqrt{[3-(-2)]^{2}+(-4-5)^{2}}=\sqrt{5^{2}+(-9)^{2}}=\sqrt{25+81}=\sqrt{106}$ units

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

Thus, $\mathrm{AB}=\mathrm{BC}=\sqrt{106}$ units

∴ ∆ABC is an isosceles triangle.

Also,

AB2 + BC2 = 106 + 106 = 212

and CA2 = 212

∴ AB2 + BC2 = CA2

So, ∆ABC is right angled at B.              (Converse of Pythagoras theorem)

Hence, the given points are the vertices of an isosceles right triangle.