Question

The coordinate of vertices of triangles are given. Identify the types of triangles $(3,5)(-1,1)(6,2)$.

Answer 1

We know that the distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Let the given vertices be $A=(3,5)$, $B=(-1,1)$ and $C=(6,2)$

We first find the distance between $A=(3,5)$ and $B=(-1,1)$ as follows:

$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(-1-3)}^2+{(1-5)}^2}=\sqrt{{(-4)}^2+{(-4)}^2}=\sqrt{16+16}=\sqrt{32}=\sqrt{4^2\times 2}$

$=4\sqrt{2}$

Similarly, the distance between $B=(-1,1)$ and $C=(6,2)$ is:

$BC=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(6-(-1\left)\right)}^2+{(2-1)}^2}=\sqrt{{(6+1)}^2+1^2}=\sqrt{7^2+1^2}=\sqrt{49+1}=\sqrt{50}$

$=\sqrt{5^2\times 2}=5\sqrt{2}$

Now, the distance between $C=(6,2)$ and $A=(3,5)$ is:

$CA=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(3-6)}^2+{(5-2)}^2}=\sqrt{{(-3)}^2+3^2}=\sqrt{9+9}=\sqrt{18}=\sqrt{3^2\times 2}=3\sqrt{2}$

We also know that if $A{B}^2+C{A}^2=B{C}^2$, then the triangle $ABC$ is a right angled triangle.

Thus consider,

$A{B}^2+C{A}^2=(4\sqrt{2}{)}^2+(3\sqrt{2}{)}^2=(16\times 2)+(9\times 2)=32+18=50=(5\sqrt{2}{)}^2=B{C}^2$

Since $A{B}^2+C{A}^2=B{C}^2$, therefore, $ABC$ is a right angled triangle.

Hence, the given vertices are the vertices of right angled triangle.