What is the area, in square units, of triangle ABC?

We need to  find the distance between

(0,6) and (-4,3)  = √[ ( -4)^2 + (6-3)^2 ] = √ [4^2 + 3^2 ]  =√[25]   = 5

Now, ACG....we could find the lengths of the other two sides and use something known as "Heron's Formula" to calculate the area, but it's messy.....here's an easier way....let the first distance we calculated be  the  base  of the triangle

We need to find the equation of the line between the first two points....

The slope is   [ 6 -3] / [ 0 - -4]  =  3/4

The equation is

y  = (3/4)(x - 0) + 6

y = (3/4)x  + 6      ....... now.... we need this form :  Ax + BY + C  = 0....so ....multiply through by  4.....

4y = 3x + 24

3x  - 4y + 24  = 0

Now the distance  between (2, -2)  and  this line will be the altitude of the triangle and  it is given by

l Ax + By + C l  /  √[ A^2 + B^2 ]       where  (x,y)  = (2, -2)   ....so we have

 l 3(2)  - 4(-2) + 24  l   / √ [ (3)^2 + (-4)^2 ]  =

l 6  + 8  + 24  l  / √25   =

38 / 5     units

So....the area  is

(1/2) base * altitude  =

(1/2) (5) * 38/ 5  =

19 units^2 

As Chris mentioned:

I calculated the distances between the points (side lengths) as  

5    7.81   and    8.25

thus p = 10.53

sqrt(10.53(10.53-5)(10.53-7.81)(10.53-8.25)) = 19.0 units^2    

                                                               (just as Chris found!)

Thank you both!

What is the area, in square units, of triangle ABC?