Question

Hey guys
Prove that the points (3,0);(6,4);(-1,3) are vertices of a right angled isosceles triangle

Answer 1

find distances between points then you get that

Answer 2

The angles of a right angled isosceles triangle is always 45°, 45° and 90°, and the sides are in the ratio 1 : 1 : √2.

Now we have to find the length of each side of the triangle, i.e., distance between each two points.

$\texttt{Distance between points (3, 0) and (6, 4) = 5} \\ \\ \\ \sqrt{(6-3)^2+(4-0)^2} \\ \\ \sqrt{3^2+4^2} \\ \\ \sqrt{9+16} \\ \\ \sqrt{25} \\ \\ \large \text{5}$

$\texttt{Distance between points (6, 4) and (-1, 3) = \mathtt{5\sqrt{2}} } \\ \\ \\ \sqrt{(6-(-1))^2+(4-3)^2} \\ \\ \sqrt{7^2+1^2} \\ \\ \sqrt{49+1} \\ \\ \sqrt{50} \\ \\ \large 5\sqrt{2}$

$\texttt{Distance between points (3, 0) and (-1, 3) = 5} \\ \\ \\ \sqrt{(3-(-1))^2+(0-3)^2} \\ \\ \sqrt{4^2+(-3)^2} \\ \\ \sqrt{16+9} \\ \\ \sqrt{25} \\ \\ \large \text{5}$

Thus the lengths of the sides of the triangle are 5 unit, 5 unit and 5√2.

From this, we get that the triangle is isosceles.

Observe the lengths of the sides. They're also in the ratio 1 : 1 : √2.

5 : 5 : 5√2  =  1 : 1 : √2

Thus it will be a right angled isosceles triangle according to the concept above.

Hence Proved!!!