Find the perimeter & area of triangle ABC, whose vertices are A(-4,-2), B(8,-2) & C(2,6). Find the length of the perpendicular from A to BC.

A(-4,-2)  B(8,-2)  C(2,6)

let
us calculate the length of the sides:

AB = sqrt[(8--4)^2 +
(-2--2)^2]= sqrt(12^2)= 12

AC= sqrt[(2--4)^2 + (6--2)^2]=
sqrt(100) = 10

BC= sqrt[(2-8)^2 + (6--2)^2]= sqrt(100)=
10

We notice that ABS is an isosceles
triangle.

The perimeter (P) = AB + AC + BC=
12+ 10 + 10 = 32

Since it is an isosceles,
where AC = BC

==> let the perpendicular line from C
on AB divides AB in midpoint

Let D be the
midpoint,

==> AD = BD = 12/2=
6

But :

AD^2 = AC^2 -
CD^2

         = 100- 36 =
64

==> The perpendicular line from C
to AB (AD) = 8

The area of the triangle =
(1/2)*AD* AB

                                       =
(1/2)*8*12= 48

The area (a) =
48