Question

Vertices od triangle are (-2,0) (2,3) (1,-3) is equilateral, scalene or isosceles also find area of triangle

Answer 1

Let, A (-2,0), B (2,3), C (1,-3)

AB=sqrt ((-2-2)^2+(0-3)^2)=5

BC=sqrt ((2-1)^2+(3+3)^2)=6.08

CA=sqrt ((1+2)^2+(-3-0)^2)=4. 24

So, triangle ABC is a scalene

Now, half Perimeter of ABC is s= (5+6. 08+4. 24)/2=7. 66

Then, area of ABC=sqrt. ((((s*(s-a).*(s-b)*(s-c))))=10. 50

Answer 2

Let A (-2,0) , B (2,3) and C (1,-3) be the vertices of ∆ABC,Then,

$\sf{AB= \sqrt{ {(2 + 2)}^{2} + {(3 - 0)}^{2} } = \sqrt{16 + 9} = \sqrt{25} = 5}$

$\sf{BC= \sqrt{ {(1 - 2)}^{2} + {( - 3 - 3)}^{2} } = \sqrt{1 + {( - 6)}^{2} } = \sqrt{37} }$

$\sf{AC = \sqrt{ {(1 + 2)}^{2} + {( - 3 - 0)}^{2} } = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} }$

Clearly, AB ≠ BC ≠ AC,Hence, ∆ABC is a scalene triangle.