# A triangle hat(ABC) has vertices of A(1,3);B(1/2,3/2);C(2,1). Verify that the triangle is isosceles and calculate the area and perimeter?

Jul 7, 2017

$\text{ Area="5/4," the Perimeter=} \sqrt{5} \left(1 + \sqrt{2}\right) .$

#### Explanation:

Let us first verify that the given points are non-collinear.

We use the following necessary and sufficient condition for

collinearity of the points :

$A \left({x}_{1} , {y}_{1}\right) , B \left({x}_{2} , {y}_{2}\right) \mathmr{and} C \left({x}_{3} , {y}_{3}\right) \text{ are collinear } \iff$

$| \left({x}_{1} , {y}_{1} , 1\right) , \left({x}_{2} , {y}_{2} , 1\right) , \left({x}_{3} , {y}_{3} , 1\right) | = 0.$

We have, $D = | \left(1 , 3 , 1\right) , \left(\frac{1}{2} , \frac{3}{2} , 1\right) , \left(2 , 1 , 1\right) | ,$

$= 1 \left(\frac{3}{2} - 1\right) - 3 \left(\frac{1}{2} - 2\right) + 1 \left(\frac{1}{2} - 3\right) ,$

$= \frac{1}{2} + \frac{9}{2} - \frac{5}{2} = \frac{5}{2.}$

Thus, the points are not collinear, and, hence, form $\Delta A B C .$

Knowing that, the Area of $\Delta A B C$ is $\frac{1}{2} | D | ,$

The Reqd. Area =$\frac{5}{4.}$

Using the Distance Formula, we have,

$A {B}^{2} = {\left(1 - \frac{1}{2}\right)}^{2} + {\left(3 - \frac{3}{2}\right)}^{2} = \frac{1}{4} + \frac{9}{4} \Rightarrow A B = \frac{\sqrt{10}}{2.}$

$B {C}^{2} = {\left(\frac{1}{2} - 2\right)}^{2} + {\left(\frac{3}{2} - 1\right)}^{1} = \frac{9}{4} + \frac{1}{4} \Rightarrow B C = \frac{\sqrt{10}}{2.}$

$A {C}^{2} = {\left(1 - 2\right)}^{2} + {\left(3 - 1\right)}^{2} = 1 + 4 \Rightarrow A C = \sqrt{5.}$

$\because , \text{ in "DeltaABC, AB=BC, :., Delta "is isisceles,}$ having

$\text{perimeter=} A B + B C + A C = \sqrt{5} + \sqrt{10} = \sqrt{5} \left(1 + \sqrt{2}\right) .$

Enjoy Maths.!