Question

The coordinate of vertices of triangles are given. Identify the types of triangles $(1,6)(3,2)(10,8)$.

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 $A=(1,6)$, $B=(3,2)$ and $C=(10,8)$

We first find the distance between $A=(1,6)$ and $B=(3,2)$ as follows:

$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(3-1)}^2+{(2-6)}^2}=\sqrt{2^2+{(-4)}^2}=\sqrt{4+16}=\sqrt{20}=\sqrt{2^2\times 5}=2\sqrt{5}$

Similarly, the distance between $B=(3,2)$ and $C=(10,8)$ is:

$BC=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(10-3)}^2+{(8-2)}^2}=\sqrt{7^2+6^2}=\sqrt{49+36}=\sqrt{85}$

Now, the distance between $C=(10,8)$ and $A=(1,6)$ is:

$CA=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(10-1)}^2+{(8-6)}^2}=\sqrt{9^2+2^2}=\sqrt{81+4}=\sqrt{85}$

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 $BC=CA=\sqrt{85}$

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