Step 1: Find the three vertices.
The triangle is made by the lines $L_1:\,y=x$, $L_2:\,x-2y+1=0$ and $L_3:\,3x-4y-1=0$. We get each vertex by solving the lines two at a time.
Step 2: Vertex A from $L_1$ and $L_2$.
Put $y=x$ into $x-2y+1=0$ to get $x-2x+1=0$, so $x=1$ and $y=1$. Hence $A(1,1)$.
Step 3: Vertex B from $L_1$ and $L_3$.
Put $y=x$ into $3x-4y-1=0$ to get $3x-4x-1=0$, so $-x=1$, giving $x=-1$ and $y=-1$. Hence $B(-1,-1)$.
Step 4: Vertex C from $L_2$ and $L_3$.
From $x-2y+1=0$ we have $x=2y-1$. Put this in $3x-4y-1=0$: $3(2y-1)-4y-1=0$, that is $6y-3-4y-1=0$, so $2y=4$, $y=2$ and $x=3$. Hence $C(3,2)$.
Step 5: Altitude from A, perpendicular to side BC.
Side $BC$ lies on $L_3$ whose slope is $\tfrac{3}{4}$. The altitude through $A$ is perpendicular, so its slope is $-\tfrac{4}{3}$. Its equation through $A(1,1)$ is $y-1=-\tfrac{4}{3}(x-1)$, i.e. $4x+3y-7=0$.
Step 6: Altitude from B, perpendicular to side AC.
Side $AC$ lies on $L_2$ whose slope is $\tfrac{1}{2}$. The altitude through $B$ has slope $-2$. Through $B(-1,-1)$: $y+1=-2(x+1)$, i.e. $2x+y+3=0$.
Step 7: Solve the two altitudes for the orthocentre.
From $2x+y+3=0$ we get $y=-2x-3$. Put into $4x+3y-7=0$: $4x+3(-2x-3)-7=0$, so $4x-6x-9-7=0$, giving $-2x=16$, $x=-8$, and $y=-2(-8)-3=13$.
So the orthocentre is the point $(-8,13)$. \[ \boxed{(-8,\,13)} \]