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

Question:

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

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(64) and C(13).

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.

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