Step 1: Use the perpendicularity of altitudes directly.
Let A(0,0), B(4,0), C(3,4), and let H(x,y) be the orthocenter. The altitude from A is perpendicular to BC, and the altitude from B is perpendicular to AC.
Step 2: Altitude from A.
\(\vec{BC} = C - B = (-1,4)\), and \(\vec{AH}=(x,y)\) must satisfy \(\vec{AH}\cdot \vec{BC}=0\):
\[-x + 4y = 0 \implies x = 4y\]
Step 3: Altitude from B.
\(\vec{AC} = C - A = (3,4)\), and \(\vec{BH}=(x-4,y)\) must satisfy \(\vec{BH}\cdot \vec{AC}=0\):
\[3(x-4) + 4y = 0 \implies 3x + 4y = 12\]
Step 4: Solve together.
Substitute \(x=4y\): \(3(4y)+4y=12 \implies 16y=12 \implies y=\frac{3}{4}\), so \(x = 4\cdot\frac{3}{4}=3\).
\[\boxed{\left(3,\frac{3}{4}\right)}\]