Prove that the points (3, 0), (6, 4) and (−1, 3)

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

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 the length of the three sides of the triangle.

$A B=\sqrt{(3-6)^{2}+(0-4)^{2}}$

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

 

$=\sqrt{9+16}$

$A B=\sqrt{25}$

$B C=\sqrt{(6+1)^{2}+(4-3)^{2}}$

$=\sqrt{(7)^{2}+(1)^{2}}$

 

$=\sqrt{49+1}$

$B C=\sqrt{50}$

$A C=\sqrt{(3+1)^{2}+(0-3)^{2}}$

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

 

$=\sqrt{16+9}$

$A C=\sqrt{25}$

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 $B C^{2}=A C^{2}+A B^{2}$

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