Question

The vertices of a triangle are (-2,0), (2,3) and(1,-3) .Is the triangle equilateral isosceles or scalene?Also find the area of the triangle

Answer 1

scalele triangle

Area will be 10.5 units

Answer 2

Answer:

Scalene triangle

Area = 10.49 square units

Step-by-step explanation:

The vertices of triangle are (-2,0), (2,3) and(1,-3)

Let Point A = (-2,0)

Point B = (2,3)

Point C = (1,-3)

Now to find sides of triangle we will use distance formula :

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

Now find the length of AB

So, A = $(x_1,y_1)= (-2,0)$

B = $(x_2,y_2)= (2,3)$

Substitute the values

$d =\sqrt{(2-(-2))^2+(3-0)^2}$

$d =\sqrt{(4)^2+(3)^2}$

$d =\sqrt{16+9}$

$d =\sqrt{25}$

$d =5$

So, AB = 5 units

Now to find BC

B = $(x_1,y_1)= (2,3)$

C = $(x_2,y_2)= (1,-3)$

Substitute the values

$d =\sqrt{(1-2)^2+(-3-3)^2}$

$d =\sqrt{(1)^2+(-6)^2}$

$d =\sqrt{1+36}$

$d =\sqrt{37}$

$d =6.08$

BC = 6.08 units

To Find AC

A = $(x_1,y_1)= (-2,0)$

C = $(x_2,y_2)= (1,-3)$

Substitute the values

$d =\sqrt{(1-(-2))^2+(-3-0)^2}$

$d =\sqrt{(3)^2+(-3)^2}$

$d =\sqrt{9+9}$

$d =\sqrt{18}$

$d =4.24$

AC = 4.24 units

So, the sides of the triangle are 5 units , 6.08 units and 4.24 units

Since all the side are of unequal length

So, the triangle is scalene triangle

Now to find area of triangle we will use heron's formula:

$Area = \sqrt{s(s-a)(s-b)(s-c)}$

Where $s = \frac{a+b+c}{2}$

a,b,c are the side lengths of triangle  

a =5

b=6.08

c = 4.24

Now substitute the values :

$s = \frac{5+6.08+4.24}{2}$

$s =7.66$

$Area = \sqrt{7.66(7.66-5)(7.66-6.08)(7.66-4.24)}$

$Area = 10.49$

Hence the area of the triangle is 10.49 square units