Question

Triangle ABC has vertices at A(-2,-3), B(6,-3), C (-1,5). Answer the following and round your answers to

Answer 1

Answer:

a) Perimeter $= {\bf 8} + \sqrt {\bf 113} + \sqrt {\bf 65}$

b)Area $={\bf 96}$

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 $(x_A,y_A) ,(x_B,y_B), (x_C,y_C)$ respectively

Now find the perimeter of the triangle ABC

The perimeter is found by first finding the three distances between the three vertices $d_AB, d_BC\ and\ d_CA$ given by

$d_{AB} = \sqrt {(x_A - x_B)^2 + (y_A - y_B)^2)}$

$d_{BC} = \sqrt {(x_B - x_C)^2 + (y_B - y_C)^2)}$

$d_{CA}= \sqrt {(x_C - x_A)^2 + (x_C - y_A)^2}$

The perimeter is given by

Perimeter $=d_{AB} + d_{BC} + d_{CA}$

now find $d_{AB} = \sqrt {(x_A - x_B)^2 + (y_A - y_B)^2)}$

               $d_{AB}= \sqrt {(-2 - 6)^2 + (-3+3 )^2)}$

               $d_{AB} = \sqrt {(-8)^2 + (0)^2)}$

               $d_{AB} = \sqrt {8^2}$

               $d_{AB}= \sqrt {64}$

               $d_{AB} = 8$

Similarly we find $d_{BC} = \sqrt {(x_B - x_C)^2 + (y_B - y_C)^2)}$

                           $d_{BC}= \sqrt {(6+1)^2 + (-3-5)^2)}$

                           $d_{BC} = \sqrt {(7)^2 + (-8)^2)}$

                           $d_{BC}= \sqrt {49 + 64}$

                           $d_{BC} = \sqrt {113}$

find $d_{CA} = \sqrt {(x_C - x_A)^2 + (x_C - y_A)^2}$

      $d_{CA} = \sqrt {(-1 +2)^2 + (5+3)^2}$

      $d_{CA} = \sqrt {(1)^2 + (8)^2}$

      $d_{CA} = \sqrt {1 + 64}$

      $d_{CA} = \sqrt {65}$

Now adding the distances we get

Perimeter $=d_{AB}+ d_{BC} + d_{CA}$

Perimeter $= 8+ \sqrt {113} + \sqrt {65}$

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:

$Area= \frac{1}{2} {\det {\left[\begin{array}{ccc}x_A&x_B&x_C\\y_A&y_B&y_C\\1&1&1\end{array}\right]}}$

where det is the determinant of the three by three matrix.

$Area=\frac{1}{2}{{\det \left[\begin{array}{ccc}-2&6&-1\\ -3& -3&5\\ 1 & 1 & 1\end{array}\right]}}$

$Area=\frac{1}{2}[-2(-3-5)-6(-3-5)-1(-3+3)+3(6+1)-3(-2+1)-5(-2-6)+1(30-3)-1(-10-3)+1(6+18)]$

$Area=\frac{1}{2}[-2(-8)-6(-8)-1(0)+3(7)-3(-1)-5(-8)+1(27)-1(-13)+1(24)]$

$Area=\frac{1}{2}[16+48+0+21+3+40+27+13+24]$

$Area=\frac{1}{2} (192)$

$Area=96$