Step 1: Use a dot product:
Let the orthocenter be $H(h,k)$. Then $AH\perp BC$ and $BH\perp AC$.
Step 2: First condition:
$\vec{BC}=(1,1)$ and $\vec{AH}=(h-1,\,k+6)$. Dot product zero: $(h-1)+(k+6)=0$, so $h+k=-5$.
Step 3: Second condition:
$\vec{AC}=(2,4)$ and $\vec{BH}=(h-2,\,k+3)$. Dot product zero: $2(h-2)+4(k+3)=0$, so $h+2k=-4$.
Step 4: Solve:
Subtract: $k=1$, then $h=-6$.
Step 5: Check with option (A):
$(-6,1)$ satisfies both equations, so (A).
Final Answer:
Both altitude conditions are satisfied by (-6, 1).
\[ \boxed{A} \]