Prove that the point (3, 0), (6, 4) and (− 1, 3) are the vertices of a right angled isosceles triangle.

We are given three points, and we have to prove that these three points are the vertices of a right-angled isosceles triangle.
Let the three vertices be 𝐴(3, 0), 𝐵(6, 4) and 𝐶(− 1, 3).
The distance between the two points gives the length of the line joining these two points. The formula can find the distance between two points,
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =$\sqrt{ {(𝑥\text{'} - 𝑥)}^2 + {(𝑦\text{'} - 𝑦)}^2}$, where (𝑥, 𝑦) and (𝑥', 𝑦') are the coordinates of the two points.
The length of 𝐴𝐵 = $\sqrt{ {(6 - 3)}^2 + {(4-0)}^2}$
⇒ 𝐴𝐵 = $\sqrt{25}$
⇒ 𝐴𝐵 = 5 𝑢𝑛𝑖𝑡𝑠
The length of 𝐵𝐶 = $\sqrt{ {(-1 - 6)}^2 + {(3-4)}^2}$
⇒ 𝐵𝐶 =$\sqrt{50}$
⇒ 𝐵𝐶 = 7. 07 𝑢𝑛𝑖𝑡𝑠
The length of 𝐶𝐴 = $\sqrt{ {(3-(-1\left)\right)}^2 + {(0-3)}^2}$

⇒ 𝐶𝐴 = $\sqrt{25}$
⇒ 𝐶𝐴 = 5 𝑢𝑛𝑖𝑡𝑠
When points 𝐴, 𝐵 and 𝐶 are joined, then ∆𝐴𝐵𝐶 is formed in which two sides are equal. The equal sides are 𝐴𝐵 and 𝐶𝐴, each equal to 5 𝑢𝑛𝑖𝑡𝑠, so ∆𝐴𝐵𝐶 is an isosceles triangle.
To check whether this triangle is right-angled or not, we can use Pythagoras’ theorem. According to Pythagoras’ theorem,
𝐵𝐶² = 𝐴𝐵²   + 𝐶𝐴²
Finding the value of RHS by putting the value of variables, we get

𝐴𝐵² + 𝐶𝐴²   = 5²   + 5²
= 25 + 25
= 50
Now, the value of LHS is
⇒ 𝐵𝐶² = 50
Thus, 𝐵𝐶² = 𝐴𝐵² + 𝐶𝐴². So, ∆𝐴𝐵𝐶 is a right-angled isosceles triangle. Hence, proved.