Show that the points A

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

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

$=\sqrt{81+25}$

$=\sqrt{106}$ units …..(1)

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

$=\sqrt{25+81}$

$=\sqrt{106}$ units …..(2)

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

$=\sqrt{16+196}$

$=\sqrt{2} 12$ units

From equations 1 and 2, we have

⇒ AB = BC

Therefore, Δ ABC is an isosceles triangle …..(3)

Also, $\mathrm{AB}^{2}=106$ units  …..(4)

$\mathrm{BC}^{2}=106$ units ...........(5)

$\mathrm{AC}^{2}=212$ units .............(6)

From equations 4, 5 and 6, we have

$\mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{AC}^{2}$

So, it satisfies the Pythagoras theorem.

Δ ABC is right angled triangle …..(7)

From 3 and 7 , we have

$\triangle \mathrm{ABC}$ is an isosceles right angled triangle.

Hence, proved.