Question

The points (1,6) and (12,9) are the two opposite vertices of a parallelogram. the other two vertices lie on the line 3y=11 x+k. then k is

Answer 1

Let (1,6) be pt-A
Let (12,9) be pt-B
So, mid-point of line AB=((12+1)/2,(9+6)/2)
                                  =(13/2,15,2)
As in a parallelogram,the diagonals bisect each-other,(13/2,15/2)lies on line-3y=11x+k
so,BTP,
3x(15/2)=11x(13/2)+k
So,k=45/2-143/2
so,k=-98/2=-49

Answer 2

The diagonal of the parallelogram has the end points (1,6) and (12,9)
The midpoint of the diagonal = $( \frac{12+1}{2}, \frac{6+9}{2}) = ( \frac{13}{2}, \frac{15}{2})$

This midpoint lies on the equation 3y = 11x + k

3(15/2) = 11(13/2) + k

45/2 = 143/2 + k

k = 45/2 - 143/2 

k = -98/2 = -49