user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x30,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  lea    D49(%rip),%rdi
  call   <puts@plt>
  lea    0x8(%rsp),%rbp
  mov    %rbp,%rsi
  lea    D130(%rip),%rbx
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D33(%rip),%rdi
  call   <puts@plt>
  lea    0x10(%rsp),%r12
  mov    %r12,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D108(%rip),%rdi
  call   <puts@plt>
  lea    0x18(%rsp),%r13
  mov    %r13,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  lea    D39(%rip),%rdi
  call   <puts@plt>
  lea    0x20(%rsp),%r14
  mov    %r14,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <calcAngularCoeff@plt>
  movq   %xmm0,%rbx
  mov    %r14,%rcx
  mov    %r13,%rdx
  mov    %r12,%rsi
  mov    %rbp,%rdi
  call   <calcIntercetta@plt>
  movapd %xmm0,%xmm1
  movsd  0x8(%rsp),%xmm2
  movsd  0x18(%rsp),%xmm0
  ucomisd %xmm0,%xmm2
  jp     L138
  movsd  D27(%rip),%xmm3
  jne    L138
L49:
  ucomisd 0x10(%rsp),%xmm3
  jp     L130
  movsd  D27(%rip),%xmm3
  jne    L130
L113:
  ucomisd 0x20(%rsp),%xmm3
  jp     L86
  je     L52
L86:
  subsd  %xmm0,%xmm2
  ucomisd D0(%rip),%xmm2
  jp     L91
  je     L74
L91:
  movq   %rbx,%xmm4
  ucomisd D27(%rip),%xmm4
  setnp  %al
  mov    $0x0,%edx
  cmovne %edx,%eax
  test   %al,%al
  je     L106
  ucomisd D0(%rip),%xmm1
  setnp  %al
  cmovne %edx,%eax
  test   %al,%al
  jne    L96
L106:
  movq   %rbx,%xmm5
  ucomisd D0(%rip),%xmm5
  setnp  %al
  mov    $0x0,%edx
  cmovne %edx,%eax
  test   %al,%al
  je     L46
  ucomisd D27(%rip),%xmm1
  setnp  %al
  cmovne %edx,%eax
  test   %al,%al
  je     L46
  lea    D15(%rip),%rdi
  call   <puts@plt>
L45:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L141
  mov    $0x0,%eax
  add    $0x30,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L138:
  pxor   %xmm3,%xmm3
  jmp    L49
L130:
  pxor   %xmm3,%xmm3
  jmp    L113
L52:
  lea    D17(%rip),%rdi
  call   <puts@plt>
  jmp    L45
L74:
  lea    D77(%rip),%rdi
  call   <puts@plt>
  jmp    L45
L96:
  lea    D81(%rip),%rdi
  call   <puts@plt>
  jmp    L45
L46:
  movq   %rbx,%xmm0
  lea    D53(%rip),%rdi
  mov    $0x2,%eax
  call   <printf@plt>
  jmp    L45
L141:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_tag: D49
data_type: string
name: disassemble_data
arguments: data_tag: D130
data_type: string
name: disassemble_data
arguments: data_tag: D33
data_type: string
name: disassemble_data
arguments: data_tag: D108
data_type: string
name: disassemble_data
arguments: data_tag: D39
data_type: string
name: disassemble_data
arguments: data_tag: D27
data_type: f32[2]
name: disassemble_data
arguments: data_tag: D0
data_type: f32[2]
name: disassemble_data
arguments: data_tag: D15
data_type: string
name: disassemble_data
arguments: data_tag: D17
data_type: string
name: disassemble_data
arguments: data_tag: D77
data_type: string
name: disassemble_data
arguments: data_tag: D81
data_type: string
name: disassemble_data
arguments: data_tag: D53
data_type: string

tool: D49: Enter the value of x1:

tool: D130: %lf

tool: D33: Enter the value of y1:

tool: D108: Enter the value of x2:

tool: D39: Enter the value of y2:

tool: {"D27": [0.0,1.875]}

tool: {"D0": [0.0,0.0]}

tool: D15: The equation of the line is constant: y = 1.000

tool: D17: The given points are equals.

tool: D77: The line is vertical.

tool: D81: The equation of the line goes through the origin: y = 1.000 * x

tool: D53: y = %lf x + %lf


assistant: ```c
int main() {
  double x1, y1, x2, y2, m = 0, q = 0;
  printf("Enter the value of x1:\n");
  scanf("%lf", &x1);
  printf("Enter the value of y1:\n");
  scanf("%lf", &y1);
  printf("Enter the value of x2:\n");
  scanf("%lf", &x2);
  printf("Enter the value of y2:\n");
  scanf("%lf", &y2);
  m = calcAngularCoeff(&x1, &y1, &x2, &y2);
  q = calcIntercetta(&x1, &y1, &x2, &y2);
  if (x1 == x2 == y1 == y2) {
    printf("The given points are equals.\n");
    return 0;
  } else if ((x1 - x2) == 0) {
    printf("The line is vertical.\n");
    return 0;
  } else if ((m == 1) && (q == 0)) {
    printf("The equation of the line goes through the origin: y = 1.000 * x\n");
    return 0;
  } else if ((m == 0) && (q == 1)) {
    printf("The equation of the line is constant: y = 1.000\n");
    return 0;
  } else {
    printf("y = %lf x + %lf\n", m, q);
    return 0;
  }
  return 0;
}
```