If the equations of the three sides of a triangle are 2x + 3y =1, 3x–2y +6 = 0 and x + y =1, then the orthocentre of the triangle lies on the line?

Solution) 
Let equation of
AB be  2x+3y-1= 0 ---(1)
BC be 3x-2y+6=0 --(2)
and
AC be  x+y-1 =0---(3)
On solving (1) and (2) ,we get,
B = (x,y) = (-16/13,15/13)
and on solving (1) and (3) ,we get,
A = (x,y) = (2,-1)
equation of BC is 3x-2y+6=0
Altitude AD is perpendicular to BC.
Therefore, equation of AD is x+y+k=0
AD is passing through A (2,-1)
Now, 
2-1+k=0
1+k=0
k=-1
Therefore, equation of AD is x+y-1=0...(4)
Altitude BE is perpendicular to AC.
Let equation of DE be x-2y = k
BE is passing through D  (-16/13,15/13)
(-16/13)-2(15/13) = k
k = -46/13
Equation of BE = x - 2y = (-46/13) ...(5)
On solving (4) and (5), we get,
the point of intersection is (-256/39, -59/39) which is the ortho centre of triangle.

Thus, the equation of orthocentre lies on the line 169x +26 y = -178




Regards!