Trying to show an expression is always zero

Question

I have a symbolic matrix

$$m = \begin{pmatrix}A1 & B1 & C1\\A2 & B2 & C2 \\ A3 & B3 & C3\end{pmatrix}$$

and I believe that Det[m] is always zero (i.e. [A3 B3 C3] is a linear combination of [A1 B1 C1] and [A2 B2 C2]).

The expressions for the A's, B's, and C's that I have are fairly complex, so by the time you multiply all the terms to get the determinant the expression is huge. I have tried substituting random numerical values for all of my symbols and the numerical result of the determinant is indeed very small (O(10^-12)), and graphically all of the lines that I've seen produced in this form intersect at the same point, so I'm quite convinced this should be zero, but I just want something to show me "yes, it is actually zero" before I proceed with this assumption in further work.

I'm not sure how to post a .nb file, so I've simplified what I have to plain text. Does anyone know how to show that det[m] = 0 without substituting numerical values for all of the symbols remaining (and obtaining very small, but unfortunately non-zero, values)?

A1 = x1*D23 + x2*D31 + x3*D12
B1 = y1*D23 + y2*D31 + y3*D12
C1 = -.5*(D12*D23*D31 + (x1^2 + y1^2)*D23 + (x2^2 + y2^2)*D31 + (x3^2 + y3^2)* D12)

A2 = x1*D24 + x2*D41 + x4*D12
B2 = y1*D24 + y2*D41 + y4*D12
C2 = -.5*(D12*D24*D41 + (x1^2 + y1^2)*D24 + (x2^2 + y2^2)*D41 + (x4^2 + y4^2)* D12)

A3 = x2*D34 + x3*D42 + x4*D23
B3 = y2*D34 + y3*D42 + y4*D23
C3 = -.5*(D23*D34*D42 + (x2^2 + y2^2)*D34 + (x3^2 + y3^2)*D42 + (x4^2 + y4^2)* D23)

m = {
     {A1, B1, C1},
     {A2, B2, C2},
     {A3, B3, C3}
    }

determinant = Det[m]

D12 = D2 - D1

D13 = D3 - D1

D14 = D4 - D1

D23 = D3 - D2

D24 = D4 - D2

D34 = D4 - D3

D21 = -D12

D31 = -D13

D32 = -D23

D41 = -D14

D42 = -D24

D43 = -D34

D1 = Sqrt[(x0 - x1)^2 + (y0 - y1)^2]

D2 = Sqrt[(x0 - x2)^2 + (y0 - y2)^2]

D3 = Sqrt[(x0 - x3)^2 + (y0 - y3)^2]

D4 = Sqrt[(x0 - x4)^2 + (y0 - y4)^2]

Random numerical substitutions

x1 = 3.4
y1 = 5.6
x2 = 7.8
y2 = 1.2
x3 = 8.3
y3 = 9.1
x4 = 7.2
y4 = 5.9
x0 = 14.08
y0 = 2.34

In[58]:= determinant

Out[58]= 7.27596*10^-12

Answer 1

I haven't seen that you want to replace more variables after the Det call. When I create your matrix completely, it can be shown that the determinant is zero

A1 = x1*D23 + x2*D31 + x3*D12;
B1 = y1*D23 + y2*D31 + y3*D12;
C1 = -1/2*(D12*D23*D31 + (x1^2 + y1^2)*D23 + (x2^2 + y2^2)*
      D31 + (x3^2 + y3^2)*D12);

A2 = x1*D24 + x2*D41 + x4*D12;
B2 = y1*D24 + y2*D41 + y4*D12;
C2 = -1/2*(D12*D24*D41 + (x1^2 + y1^2)*D24 + (x2^2 + y2^2)*
      D41 + (x4^2 + y4^2)*D12);

A3 = x2*D34 + x3*D42 + x4*D23;
B3 = y2*D34 + y3*D42 + y4*D23;
C3 = -1/2*(D23*D34*D42 + (x2^2 + y2^2)*D34 + (x3^2 + y3^2)*
      D42 + (x4^2 + y4^2)*D23);

m = ({{A1, B1, C1}, {A2, B2, C2}, {A3, B3, C3}}) //. {
   D12 -> D2 - D1, D13 -> D3 - D1, D14 -> D4 - D1, D23 -> D3 - D2,
   D24 -> D4 - D2, D34 -> D4 - D3, D21 -> -D12, D31 -> -D13, 
   D32 -> -D23, D41 -> -D14, D42 -> -D24, D43 -> -D34, 
   D1 -> Sqrt[(x0 - x1)^2 + (y0 - y1)^2],
   D2 -> Sqrt[(x0 - x2)^2 + (y0 - y2)^2],
   D3 -> Sqrt[(x0 - x3)^2 + (y0 - y3)^2],
   D4 -> Sqrt[(x0 - x4)^2 + (y0 - y4)^2]
   }

Simplify[Det[m] == 0]

give then True.