Tiziana, there isn't a single formula for finding the area of triangles and in this case, you cannot simply use the formula 1/2*base*height. You would need complicated formulas for this 1. Since it would be pretty difficult to explain you the more complicated but faster formulas, I would use a less complicated but a fairly long formula. Please take some time to read it carefully and everything would go into your head. Here it goes..
First, since A,B and C are points on a co-ordinate axis and you do not know the sides of the triangle, you will first have to find them using the distance formula. If don't know what that is, then read on..
Acc to distance formula,
if A and B r points on the co-ordinate axis, such that
A=(x1,y1) and B=(x2,y2) then
AB^2 = (x2-x1)^2 + (y2-y1)^2
therefore, AB = [ (x2-x1)^2 + (y2-y1)^2]^1/2 (^1/2 means its sq root)
So now by applying this formula in the points which you gave which are A=(-1,1) , B=(4,3) , C=(0,13)
AB=[(-1-4)^2 + (1-3)^2]^1/2
BC=[(4-0)^2 + (3-13)^2]^1/2
BC=[16 + 100]^1/2
AC=[(-1-0)^2 + (1-13)^2]^1/2
Now, when you have all the sides, to find the area of the triangle we use the Heron's Formula which states-
Area of the triangle=[s(s-a)(s-b)(s-c)]^1/2
where a,b,c are sides of the triangle and s=(a+b+c)/2
In this case,
a=√29, b=√116, c=√145
s=(√29 + √116 + √145)/2
=(√29 + √4*29 + √5*29)/2
=(√29 + 2*√29 + √5*√29)/2
So now by put these values in the Heron's Formula and find the answer. I could have found you the answer but this post is getting just too long. You just need a calc for this part.
Hope this helps ;)