Question

The coordinate of vertices of triangles are given. Identify the types of triangles $(2,1)(10,1)(6,9)$.

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 $A=(2,1)$, $B=(10,1)$ and $C=(6,9)$

We first find the distance between $A=(2,1)$ and $B=(10,1)$ as follows:

$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(10-2)}^2+{(1-1)}^2}=\sqrt{8^2+0^2}=\sqrt{64}=\sqrt{8^2}=8$

Similarly, the distance between $B=(10,1)$ and $C=(6,9)$ is:

$BC=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(6-10)}^2+{(9-1)}^2}=\sqrt{{(-4)}^2+8^2}=\sqrt{16+64}=\sqrt{80}=\sqrt{4^2\times 5}=4\sqrt{5}$

Now, the distance between $C=(6,9)$ and $A=(2,1)$ is:

$CA=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(6-2)}^2+{(9-1)}^2}=\sqrt{4^2+8^2}=\sqrt{16+64}=\sqrt{80}=\sqrt{4^2\times 5}=4\sqrt{5}$

We also know that If any two sides have equal side lengths, then the triangle is isosceles.

Here, since the lengths of the two sides are equal that is $BC=CA=4\sqrt{5}$

Hence, the given vertices are the vertices of an isosceles triangle.