Question

Find the co ordinates of the orthocentre of the triangle formed by the straight lines x+y_1=0,x+2y_4=0 and x+3y_9=0

Answer 1

Orthocentre is the intersection of altitudes of triangle.
so, first of all we have to find at least two altitudes of triangle.
Let equation of side AB : x + y - 1 = 0
equation of side BC : x + 2y - 4 = 0
equation of side CA : x + 3y - 9 = 0

Altitude on BC :
slope of BC = -1/2
slope of altitude on BC = -1/slope of BC = 2
intersection of AB and CA
x + y - 1 = 0
x + 3y - 9 = 0
_____________
2y = 8 ⇒y = 4
and x = -3
Hence, altitude on BC passing through (-3,4)
So, equation of altitude on BC : (y - 4) -2(x + 3) = 0
y - 2x -10 = 0 ------(1)

Similarly altitude on CA :-
slope of CA = -1/slope of CA = 3
intersection of AB and BC
x + y - 1 = 0
x + 2y - 4 = 0
____________
y = 3 and x = -2
∴equation of altitude on CA :(y -2) -3(x + 2) = 0
y - 3x - 8 = 0 --------(2)

Solve equation (1) and (2),
x = 2 and y = 14

Hence, orthocentre ≡(2,14)