The only line capable of being perpendicular to $3x-4y=0$ is $4x+3y=4$
The point opposite to this side is $\left( \frac 32, \frac 12 \right)$
Clearly, $3x-4y=0$ does not pass through this particular point, and hence it cannot pass through the orthocentre.
What’s wrong with my proof?
On
Refer to the graph:
$\hspace{2cm}$
The altitude to the red line $y=\frac13x$ passing through $(2,-1)$ is: $$y=-3x+5$$ The altitude to the blue line $y=-\frac43x+\frac53$ passing through $(1.5,0.5)$ is: $$y=\frac34x-\frac58$$ The altitude to the green line $y=-3x+5$ passing through $(1,\frac13)$ is: $$y=\frac13x$$ The three altitudes intersect at $(1.5,0.5)$.
Clearly, the line $3x-4y=0$ does not pass through this point. And it must be a typo in the source.
Actually $4x+3y=k$ will be perpendicular to $3x-4y=0$ for any $k$.
The first and third lines $x-3y=0$ and $3x+y=5$ are perpendicular to each other and thus their intersection $(3/2,1/2)$ is the orthocenter of the triangle. But this does not match $3x-4y=0$ so the claim as written is false.
Do not dismiss the possibility of a typographical error. If the third line is $3x+y=\color{#0055ff}{0}$, putting the orthocenter at the origin, the proof works.