Find the area of the triangle whose vertices (on cartesian graphs) are (-1,5) , (-2,-3) & (10,1)

Answer 1

I don't know at what of math this is from.
There are several ways to do this, I will show you two ways:

1. pick one line segment to be the base of your triangle,
I picked the line from (-1,5) to (-2,-3)
slope = (-3-5)/(-2+1) = -8/-1 = 8
equation:
y-5 = 8(x+1)
8x + 8 = y-5
8x - y + 13 = 0

distance form an external point (a,b) to a line
Ax + By + C = 0 is |aA + bB + C|/√(A^2 + B^2)

so from (10,1) to our line
= |10(8) + 1(-1) + 13|/√(64+1) = 92/√65

length of base = √( (-3-5)^2 + (-2+1)^2) = √65

area of triangle = (1/2) base x height
= (1/2)(√65)(92/√65) = 46

2. The simplest way involves the cross-product of vectors

line up your 3 points in a column, repeating the point you started with at the end

-1 5
-2 -3
10 1
-1 5

Area = (1/2) | sum of downproducts - sum of upproducts |
= (1/2) |(3 - 2 + 50) - (-10 -30 -1) |
=(1/2) | 51 + 41|
= (1/2)(92) = 46

BTW, this works for all convex polygons if you know the points, make sure you list them in sequence, repeating the one you started with