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

#### Solution 1

Let A(3, 0), B(6, 4) and C(–1, 3) be the given points

Now,

AB= `sqrt((6-3)^2+(4-0)^2)=sqrt(3^2+4^2)=sqrt(9+6)=sqrt25`

BC= `sqrt((-1-6)^2+(3-4)^2)=sqrt((-7)^2+(-1)^2)=sqrt(49+1)=sqrt50`

AC= `sqrt((-1-3)^2+(3-0)^2)=sqrt((-4)^2+3^2)=sqrt(16+9)=sqrt25`

∴ AB = AC

AB^{2} =`(sqrt25)=25`

BC^{2}= `(sqrt50)=50`

AC^{2}= `(sqrt25)=25`

∴ AB^{2} + AC^{2} = BC^{2}

Thus, ΔABC is a right-angled isosceles triangle.

#### Solution 2

The distance *d* between two points `(x_1,y_1)` and `(x_2, y_2)` is given by the formula

`d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`

In an isosceles triangle there are two sides which are equal in length.

Here the three points are *A*(3*,* 0)*, B*(6*, *4) and *C*(*−*1*, *3).

Let us check th`e length of the three sides of the triangle.

`AB = sqrt((3 - 6)^2 + (0 - 4)^2)`

`= sqrt((-3)^2 + (-4)^2)`

`= sqrt(9 + 16)`

`AB = sqrt(25)`

`BC = sqrt((6 + 1)^2 + (4 - 3)^2)`

`= sqrt((7)^2 + (1)^2)`

`= sqrt(49 + 1)`

`BC = sqrt50`

`AC = sqrt((3 + 1)^2 + (0 - 3)^2)`

`= sqrt((4)^2 + (-3)^2)`

`= sqrt(16 - 9)`

`AC = sqrt25`

Here, we see that two sides of the triangle are equal. So the triangle formed should be an isosceles triangle.

We can also observe that `BC^2 = AC^2 + AB^2`

Hence proved that the triangle formed by the three given points is an isosceles triangle.