Chapter E, Problem 13.10EX

To determine

To identify the type of the triangle, whose vertices are $(1,-3),(6,2)$ and $(0,4)$ .

Answer to Problem 13.10EX

Option (a)

Explanation of Solution

Given information: Vertices of the triangle are: $(1,-3),(6,2)$ and $(0,4)$ .

Concept used:

The distance between the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is

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

Let ABC be the triangle with vertices as $A(1,-3),B(6,2)$ and $C(0,4)$ .

Using distance formula,

  $\begin{array}{l}AB=\sqrt{25+25}=\sqrt{50}=5\sqrt{2};\\ BC=\sqrt{36+4}=\sqrt{40}=2\sqrt{10};\\ AC=\sqrt{1+49}=\sqrt{50}=5\sqrt{2}\end{array}$

Here, AB = AC

Hence, the triangle is an isosceles triangle.