05AB1E, 16 15 16 bytes

Saved a byte thanks to Adnan

FðN×…|ÿ\}Dg'-×»?

Try it online!

Explanation

F       }         # for N in range [0 ... input-1]
 ðN×              # push <space> repeated N times
    …|ÿ\          # to the middle of the string "|\"
         Dg       # get length of last string pushed
           '-×    # repeat "-" that many times
              »   # join strings by newline
               ?  # print without newline

C, 58

i;f(n){for(i=2*n;~i--;printf(i<n?"-":"|%*c\n",2*n-i,92));}

--

Thanks to @Steadybox who's comments on this answer helped me shave a few bytes in my above solution

C#, 93 bytes

n=>{var s=n>0?new string('-',n+1):"";while(n-->0)s="|"+new string(' ',n)+"\\\n"+s;return s;};

Anonymous function which returns the ASCII triangle as a string.

Full program with ungolfed, commented function and test cases:

using System;

class ASCIITriangles
{
    static void Main()
    {
      Func<int, string> f =
      n =>
      {
          // creates the triangle's bottom, made of dashes
          // or an empty string if n == 0
          var s = n > 0 ? new string('-', n + 1) : "";

          // a bottom to top process
          while ( n-- > 0)
          // that creates each precedent line
            s = "|" + new string(' ', n) + "\\\n" + s;

          // and returns the resulting ASCII art
          return s;
      };

      // test cases:
      Console.WriteLine(f(4));
      Console.WriteLine(f(0));
      Console.WriteLine(f(1));
    }
}

V, 18 17 bytes

Àñé\é|hñÀñÙá ñÒ-x

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Explanation

We first have a loop to check if the argument is 0. If so, the code below executes (|\ is written). Otherwise, nothing is written and the buffer is empty.

Àñ     ñ            " Argument times do:
  é\é|              " Write |\
      h             " Exit loop by creating a breaking error

Now that we got the top of the triangle, we need to create its body.

Àñ   ñ              " Argument times do:
  Ù                 " Duplicate line, the cursor comes down
   à<SPACE>         " Append a space

Now we got one extra line at the bottom of the buffer. This has to be replaced with -s.

Ó-                  " Replace every character with a -
   x                " Delete the extra '-'

This answer would be shorter if we could whatever we want for input 0

V, 14 13 bytes

é\é|ÀñÙá ñÒ-x

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CJam, 24 22 21 bytes

Saved 1 byte thanks to Martin Ender

ri_{S*'|\'\N}%\_g+'-*

Try it online!

Explanation

ri                     e# Take an integer from input
  _                    e# Duplicate it
   {                   e# Map the following to the range from 0 to input-1
    S*                 e#   Put that many spaces
      '|               e#   Put a pipe
        \              e#   Swap the spaces and the pipe
         '\            e#   Put a backslash
           N           e#   Put a newline
            }%         e# (end of map block)
              \        e# Swap the top two stack elements (bring input to the top)
               _g+     e# Add the input's signum to itself. Effectively this increments any 
                       e#  non-zero number and leaves zero as zero.
                  '-*  e# Put that many dashes

Python 2, 69 bytes

lambda x:'\n'.join(['|'+' '*n+'\\'for n in range(x)]+['-'*-~x*(x>0)])

Try it online!

C 101 93 bytes

f(n){i;for(i=0;i<n;i++){printf("%c%*c",124,i+1,92);puts("\n");}for(i=0;i<n;i++)printf("_");}

Ungolfed version

void f(int n)
{
  int i;

  for(i=0;i<n;i++)
  {    
     printf("%c%*c",124,i+1,92);
     puts("\n");    
  }
  for(i=0;i<=n;i++)
    printf("-");

}

Javascript (ES6), 97 85 81 75 bytes

n=>(g=(n,s)=>n?g(--n,`|${" ".repeat(n)}\\\n`+s):s)(n,"")+"-".repeat(n&&n+1)

Turns out I wasn't using nearly enough recursion

f=n=>(g=(n,s)=>n?g(--n,`|${" ".repeat(n)}\\\n`+s):s)(n,"")+"-".repeat(n&&n+1)

console.log(f(0))
console.log(f(1))
console.log(f(2))
console.log(f(3))
console.log(f(4))

Pyke, 18 17 bytes

I Fd*\|R\\s)Qh\-*

Try it here!

SmileBASIC, 51 bytes

INPUT N
FOR I=0TO N-1?"|";" "*I;"\
NEXT?"-"*(N+!!N)

PowerShell, 51 67 bytes

param($n)if($n){1..$n|%{"|"+" "*--$_+"\"};write-host -n ('-'*++$n)}

Try it online!

(Byte increase to account for no trailing newline)

Takes input $n and verifies it is non-zero. Then loops to construct the triangle, and finishes with a line of -. Implicit Write-Output happens at program completion.

Python2, 73 bytes

n=input()
w=0
exec'print"|"+" "*w+"\\\\"+("\\n"+"-"*-~n)*(w>n-2);w+=1;'*n

A full program. I also tried string interpolation for the last line, but it turned out be a couple bytes longer :/

exec'print"|%s\\\\%s"%(" "*w,("\\n"+"-"*-~n)*(w>n-2));w+=1;'*n

Another solution at 73 bytes:

n=j=input()
exec'print"|"+" "*(n-j)+"\\\\"+("\\n"+"-"*-~n)*(j<2);j-=1;'*n

Test cases

0:

1:
|\
--

2:
|\
| \
---

3:
|\
| \
|  \
----

6:
|\
| \
|  \
|   \
|    \
|     \
-------

MATL, 19 bytes

?'\|- '2GXyYc!3Yc!)

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?         % Implicit input. If non-zero
  '\|- '  %   Push this string
  2       %   Push 2
  G       %   Push input
  Xy      %   Identity matrix of that size
  Yc      %   Prepend a column of 2's to that matrix
  !       %   Transpose
  3       %   Push 3
  Yc      %   Postpend a column of 3's to the matrix
  !       %   Transpose
  )       %   Index into string
          % Implicit end. Implicit display

QBIC, 41 bytes

:~a>0|[a|?@|`+space$(b-1)+@\`][a+1|Z=Z+@-

Explanation

:~a>0|  Gets a, and checks if a > 0
        If it isn't the program quits without printing anything
[a|     For b=1; b <= a; b++
?@|`+   Print "|"
space$  and a number of spaces
(b-1)   euqal to our current 1-based line - 1
+@\`    and a "\"
]       NEXT
[a+1|   FOR c=1; c <= a+1; c++
Z=Z+@-  Add a dash to Z$
        Z$ gets printed implicitly at the end of the program, if it holds anything
        The last string literal, IF and second FOR loop are closed implicitly.

R, 101 bytes

for(i in 1:(n=scan())){stopifnot(n>0);cat("|",rep(" ",i-1),"\\\n",sep="")};cat("-",rep("-",n),sep="")

This code complies with the n=0 test-case if you only consider STDOUT !
Indeed, the stopifnot(n>0) part stops the script execution, displays nothing to STDOUT but writes Error: n > 0 is not TRUE to SDTERR.

Ungolfed :

for(i in 1:(n=scan()))
    {
    stopifnot(n>0)
    cat("|", rep(" ", i-1), "\\\n", sep = "")
    }

cat("-", rep("-", n), sep = "")

Python 2, 62 bytes

n=input();s='\\'
exec"print'|'+s;s=' '+s;"*n
if n:print'-'*-~n

Try it online!

Prints line by line, each time adding another space before the backslash. If a function that doesn't print would be allowed, that would likely be shorter.

JavaScript (ES6), 71 bytes

f=
n=>console.log(' '.repeat(n).replace(/./g,'|$`\\\n')+'-'.repeat(n+!!n))

<form onsubmit=f(+i.value);return!true><input id=i type=number><input type=submit value=Go!>

Outputs to the console. Save 6 bytes if printing to the SpiderMonkey JavaScript shell is acceptable. Save 13 bytes if returning the output is acceptable.

Javascript 101(Full Program), 94(Function Output), 79(Return) bytes

Full Program

Will not run in Chrome (as process doesn't exist apparently)
Will not run in TIO (as prompt apparently isn't allowed)

x=prompt();s='';for(i=0;i<x;i++)s+='|'+' '.repeat(i)+`\\
`;process.stdout.write(s+'-'.repeat(x&&x+1))

Function with EXACT print

x=>{s='';for(i=0;i<x;)s+='|'+' '.repeat(i++)+`\\
`;process.stdout.write(s+'-'.repeat(x&&x+1))}

Try it Online

Function with return string

x=>{s='';for(i=0;i<x;)s+='|'+' '.repeat(i++)+`\\
`;return s+'-'.repeat(x&&x+1)}

Try it Online

Repeating characters in Javascript is dumb and so is suppressing newlines on output

Pyth, 23 bytes

Iq0Q.q)VQ++\|*dN\\)*\-h

Test suite available online.

Explanation

Iq0Q.q)VQ++\|*dN\\)*\-h
   Q    Q              Q  Q implicitly appended to program, initializes to eval(input)
              d           d initializes to ' ' (space)
Iq0Q  )                   If 0 == eval(input):
    .q                     Exit the program
       VQ         )       For N in range(0, eval(input)):
             *dN           Repeat space N times
          +\|              Prepend |
         +      \\         Append \
                           Implicitly print on new line
                   *\-hQ  Repeat - 1 + eval(input) times
                          Implicitly print on new line

Python 2, 82 bytes

def f(i,c=0):
 if c<i:print'|'+' '*c+'\\';f(i,c+1)
 print'-'*((c+1,c)[c<1]);exit()

Try it online!

Longer that the other Python answers but a recursive function just to be different.

It feels wasteful using two print statements but I can't find a shorter way round it. Also the exit() wastes 7 to stop it printing decreasing number of - under the triangle.

Javascript (ES6), 84 bytes

Creates an anonymous function that presents the result on an alert box. No output is present if 0.

i=>{for(X=i,a='';i--;)a='|'+(' '.repeat(i))+'\\\n'+a;X&&alert(a+('-'.repeat(++X)))}

If printing is really required (137 bytes):

i=>for(X=i,d=document.body;i--;)d.innerHTML='|'+('&nbsp;'.repeat(i))+'\\<br>'+d.innerHTML;X&&window.print(d.innerHTML+=('-'.repeat(++X)))

Changes the HTML code and prints the page into a paper sheet. The page stays white if 0.

Clojure, 104 bytes

(fn[h](let[r #(apply str(repeat % %2))](doseq[n(range h)](println(str\|(r n\ )\\)))(print(r(inc h)\-))))

At least I should have lots of room to golf from here!

Loops over the range from 0 to h-1, printing a \|, then h-many spaces, then a \\. Finally, at the end, it prints h+1 many dashes.

(defn triangle [hypot-n]
  (let [r #(apply str (repeat % %2))] ; Create a shortcut string repeat function, since I need it twice.
    (doseq [line-n (range hypot-n)] ; Loop over the range 0 to hypot...
      (println (str \| (r line-n \ ) \\))) ; Printing a bar, then line-n many spaces, then a slash
    (print (r (inc hypot-n) \-)))) ; Then print the trailing dashes

Batch, 110 bytes

@set s=
@for /l %%i in (1,1,%1)do @call:c
@if %1 gtr 0 echo -%s: =-%
@exit/b
:c
@echo ^|%s%\
@set s= %s%

Special-casing 0 is no hardship as %s: =-% would fail anyway.

Common Lisp, 89 bytes

Creates an anonymous function that takes the n input and prints the triangle to *standard-output* (stdout, by default).

Golfed

(lambda(n)(when(< 0 n)(dotimes(i n)(format t"|~v@t\\~%"i))(format t"~v,,,v<~>"(1+ n)#\-)))

Ungolfed

(lambda (n)
  (when (< 0 n)
    (dotimes (i n)
      (format t "|~v@t\\~%" i))
    (format t "~v,,,v<~>" (1+ n) #\-)))

I'm sure I could make this shorter somehow.

Befunge-98, 68 bytes

&:!#@_000pv<
:kg00 ',|'<|`g00:p00+1g00,a,\'$,kg00
00:k+1g00-'<@,k+1g

Haskell, 82 bytes

f 1=["\\\n"]
f n|t<-f$n-1=t++[' ':last t]
g 0=""
g n=(f n>>=('|':))++([0..n]>>"-")

Try it online! Usage:

Prelude> g 4
"|\\\n| \\\n|  \\\n|   \\\n-----"

Or more nicely:

Prelude> putStr $ g 4
|\
| \
|  \
|   \
-----

JavaScript, 66 bytes

f=(x,i=0)=>i<x?`|${" ".repeat(i)}\\\n`+f(x,i+1):"-".repeat(x&&i+1)

Mathematica, 57 bytes

Array[{"|"," "~Table~#,"\\
"}&,#,0]<>Table["-",#+Sign@#]&

Unnamed function taking a nonnegative integer as input and returning a string-with-newlines. It deals with the special case of input 0 by noting that the number of dashes is equal to the input plus the sign of the input.

Underload, 58 bytes

:(!(:((|)S:( )~^S(\
)S(:)~*(*)*)~^(!())~^(-):S~^S))~^()~^^

Assumes the input is a Church numeral on the stack.

Explanation:

This is the main subprogram:

:((|)S:( )~^S(\
)S(:)~*(*)*)~^(!())~^(-):S~^S

It runs this code n times:

(|)S:( )~^S(\
)S(:)~*(*)*

Stack trace of that code (with 3 on the stack):

       (::**)
(|)    (::**)(|)
S      (::**)          ; printed
:      (::**)(::**)
( )    (::**)(::**)( )
~      (::**)( )(::**)
^      (::**)( )
 :     (::**)( )( )
 :     (::**)( )( )( )
 *     (::**)( )(  )
 *     (::**)(   )
S      (::**)          ; printed
(\ )   (::**)(\ )      ; the space is a newline
S      (::**)          ; printed
(:)    (::**)(:)
~      (:)(::**)
*      (:::**)
(*)    (:::**)(*)
*      (:::***)

All this does is prints the line with counter repeated times and then increments counter. So, running this n times with counter starting at 0 (!()) would print the required lines. The only thing left to do is to print the dashes at the bottom, which is just repetition.

The wrapper around the main subprogram just checks if the input was zero and do nothing if so.

8th, 94 bytes

: f 0; dup ( "|" . ( "\x20" . ) swap times "\x5c" . cr ) 0 rot n:1- loop n:1+ "-" swap s:* . ;

Explanation

The word f require an integer input

: f \ n -- 
   0;                      \ Exit if input equals to 0
   dup                     \ Duplicate input on the stack
   ( "|" .                 \ Print a piece of the vertical side
   ( "\x20" . ) swap times \ Print space between sides
   "\x5c" . cr )           \ Print a piece of the slanted side
   0 rot n:1- loop         \ Repeat the above operation for the size required 
   n:1+ "-" swap s:* .     \ Print the bottom side
;

Usage

4 f
|\
| \
|  \
|   \
-----