find the coordinates of the orthocenter of the triangle formed by the straight lines x+y-1=0, x+2y-4=0, x+3y-9=0

let ABC be the triangle formed by the given lines.

ortho-center of any triangle is the intersection point of the altitudes.
let the coordinates of the ortho-centre be O(h,k).
the coordinates of the point B are the intersection of line AB and BC.
$$\begin{gathered} x+y-1=0 \\ x+3y-9=0 \\ -2y+8=0 \\ y=\frac{8}{2}=4 \\ x+4-1=0 \\ x+3=0\Rightarrow x=-3 \end{gathered}$$
the coordinates of B are (-3,4).
the coordinates of the point C are the intersection of the line AC and BC.
$$\begin{gathered} x+2y-4=0 \\ x+3y-9=0 \\ -y+5=0\Rightarrow y=5 \\ x+2*5-4=0 \\ x+10-4=0 \\ x+6=0\Rightarrow x=-6 \\ the coordinates of C are (-6,5) \end{gathered}$$
since $OB\perp AC$
slope of OB * slope of AC = -1
$$\begin{gathered} \frac{k-4}{h+3}*\left(-\frac{1}{2}\right)=-1 \\ k-4=2*(h+3) \\ k-4=2h+6 \\ 2h-k+10=0 ..........\left(1\right) \end{gathered}$$
similarly $OC\perp AB$
slope of OC * slope of AB = -1
$$\begin{gathered} \frac{k-5}{h+6}*(-1)=-1 \\ k-5=h+6 \\ h-k+11=0 .......\left(2\right) \end{gathered}$$
solving eq(1) and eq(2):
$$\begin{gathered} 2h-k+10=0 \\ h-k+11=0 \\ h-1=0\Rightarrow h=1 \\ 2*1-k+10=0 \\ -k+12=0\Rightarrow k=12 \\ the coordinates of the ortho-center O are (1,12) \end{gathered}$$

hope this helps you