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.
the coordinates of B are (-3,4).
the coordinates of the point C are the intersection of the line AC and BC.
x+2y-4=0x+3y-9=0-y+5=0y=5x+2*5-4=0x+10-4=0x+6=0x=-6the coordinates of C are (-6,5)
since OBAC
slope of OB * slope of AC = -1
k-4h+3*-12=-1k-4=2*(h+3)k-4=2h+62h-k+10=0 ..........(1)
similarly OCAB
slope of OC * slope of AB = -1
k-5h+6*(-1)=-1k-5=h+6h-k+11=0 .......(2)
solving eq(1) and eq(2):
2h-k+10=0h-k+11=0h-1=0h=12*1-k+10=0-k+12=0k=12the coordinates of the ortho-center O are (1,12)

hope this helps you

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