Respuesta :
Answer:
a) Perimeter [tex]= {\bf 8} + \sqrt {\bf 113} + \sqrt {\bf 65} [/tex]
b)Area [tex]={\bf 96} [/tex]
Step-by-step explanation:
Given ABC is a triangle with vertices at A(-2,-3), B(6,-3) and C(-1,5)
The vertices A(-2,-3), B(6,-3) and C(-1,5) are represented by [tex](x_A,y_A) ,(x_B,y_B), (x_C,y_C)[/tex] respectively
Now find the perimeter of the triangle ABC
The perimeter is found by first finding the three distances between the three vertices [tex]d_AB, d_BC\ and\ d_CA[/tex] given by
[tex]d_{AB} = \sqrt {(x_A - x_B)^2 + (y_A - y_B)^2)}[/tex]
[tex]d_{BC} = \sqrt {(x_B - x_C)^2 + (y_B - y_C)^2)}[/tex]
[tex]d_{CA}= \sqrt {(x_C - x_A)^2 + (x_C - y_A)^2}[/tex]
The perimeter is given by
Perimeter [tex]=d_{AB} + d_{BC} + d_{CA}[/tex]
now find [tex]d_{AB} = \sqrt {(x_A - x_B)^2 + (y_A - y_B)^2)}[/tex]
[tex]d_{AB}= \sqrt {(-2 - 6)^2 + (-3+3 )^2)}[/tex]
[tex]d_{AB} = \sqrt {(-8)^2 + (0)^2)}[/tex]
[tex]d_{AB} = \sqrt {8^2}[/tex]
[tex]d_{AB}= \sqrt {64}[/tex]
[tex]d_{AB} = 8[/tex]
Similarly we find [tex]d_{BC} = \sqrt {(x_B - x_C)^2 + (y_B - y_C)^2)}[/tex]
[tex]d_{BC}= \sqrt {(6+1)^2 + (-3-5)^2)}[/tex]
[tex]d_{BC} = \sqrt {(7)^2 + (-8)^2)}[/tex]
[tex]d_{BC}= \sqrt {49 + 64}[/tex]
[tex]d_{BC} = \sqrt {113}[/tex]
find [tex]d_{CA} = \sqrt {(x_C - x_A)^2 + (x_C - y_A)^2}[/tex]
[tex]d_{CA} = \sqrt {(-1 +2)^2 + (5+3)^2}[/tex]
[tex]d_{CA} = \sqrt {(1)^2 + (8)^2}[/tex]
[tex]d_{CA} = \sqrt {1 + 64}[/tex]
[tex]d_{CA} = \sqrt {65}[/tex]
Now adding the distances we get
Perimeter [tex]=d_{AB}+ d_{BC} + d_{CA}[/tex]
Perimeter [tex]= 8+ \sqrt {113} + \sqrt {65}[/tex]
b) Area of the given triangle ABC
The formula for the area of the triangle defined by the three vertices A, B and C is given by:
[tex]Area= \frac{1}{2} {\det {\left[\begin{array}{ccc}x_A&x_B&x_C\\y_A&y_B&y_C\\1&1&1\end{array}\right]}}[/tex]
where det is the determinant of the three by three matrix.
[tex]Area=\frac{1}{2}{{\det \left[\begin{array}{ccc}-2&6&-1\\ -3& -3&5\\ 1 & 1 & 1\end{array}\right]}}[/tex]
[tex]Area=\frac{1}{2}[-2(-3-5)-6(-3-5)-1(-3+3)+3(6+1)-3(-2+1)-5(-2-6)+1(30-3)-1(-10-3)+1(6+18)] [/tex]
[tex]Area=\frac{1}{2}[-2(-8)-6(-8)-1(0)+3(7)-3(-1)-5(-8)+1(27)-1(-13)+1(24)] [/tex]
[tex]Area=\frac{1}{2}[16+48+0+21+3+40+27+13+24][/tex]
[tex]Area=\frac{1}{2} (192)[/tex]
[tex]Area=96[/tex]
