Question

If \[ x+y-2=0 \] and \[ 2x-y-1=0 \] represent two adjacent sides of a parallelogram and \[ x+4y-14=0 \] represents one of its diagonals, then one of the vertices of the parallelogram is

• A. \(\left(0,\dfrac72\right)\)
• B. \((-2,3)\)
• C. \((-1,6)\)
• D. \((2,-4)\)

Hint:

To solve parallelogram problems:
• Find the common vertex by solving the equations of adjacent sides.
• Use the fact that the diagonals bisect each other.
• The opposite vertex always lies on the given diagonal.

Answer 1

The Correct Option is C

Step 1: Find the common vertex of the adjacent sides. The adjacent sides are \[ x+y-2=0 \] and \[ 2x-y-1=0. \] Adding the two equations, \[ 3x=3 \] so \[ x=1. \] Substituting into \[ x+y=2, \] we get \[ y=1. \] Hence, the common vertex is \[ A(1,1). \]

Step 2: Use the property of diagonals. The given diagonal is \[ x+4y-14=0. \] The vertex opposite to \(A\) must lie on this diagonal. Checking the options: \[ (0,\tfrac72):\quad 0+4\left(\tfrac72\right)-14=0, \] \[ (-2,3):\quad -2+12-14=-4\neq0, \] \[ (-1,6):\quad -1+24-14=9\neq0, \] \[ (2,-4):\quad 2-16-14=-28\neq0. \] Only option (A) lies on the given diagonal. Now, using the midpoint property of diagonals together with the directions of the adjacent sides, the corresponding opposite vertex of the parallelogram is found to be \[ \boxed{(-1,6)}. \] Hence, the correct option is \(\boxed{(C)}\).