2016/079) The lines $x^2-3xy+2y^2=0$ are shifted parallel to themselves so that their point of intersection comes to (1,1). find the combined equation of lines in the new position.

we have $x^2-3xy+y^2=0$ represent two lines $(x-y)(x - 2y)= 0$ giving $(x=y=0)$. if we need point of
 intersection 1,1
then we need to replace x by x-1 and y by y-1 in given equation giving
$(x-1)^2 -3(x-1)(y-1) + 2(y-1)^2 = 0$
or $x^2 - 2x +1 -3xy + 3x + 3y -3 + 2y^2-4y + 2= 0$
or $x^2+2y^2 - 3xy + x - y= 0$