Question

Two diagonal sides of a parallelogram \(ABCD\) are \[ x-y+1=0 \] and \[ 3x-y=9. \] If one of its diagonals is \[ 7x-5y+3=0, \] then the equation of the other diagonal is

• A. \(x-3y+13=0\)
• B. \(5x+7y=13\)
• C. \(x+y-2=0\)
• D. \(x+3y=4\)

Hint:

The diagonals of a parallelogram bisect each other. Find the midpoint of the known diagonal first, then determine the equation of the other diagonal passing through the same midpoint.

Answer 1

The Correct Option is A

Step 1: Find the intersection of the given sides. The adjacent sides are \[ x-y+1=0, \] and \[ 3x-y=9. \] Solving, \[ y=x+1, \] and \[ 3x-(x+1)=9, \] \[ 2x=10, \] \[ x=5,\qquad y=6. \] Hence, \[ A=(5,6). \]

Step 2: Find the opposite vertex on the given diagonal. The given diagonal is \[ 7x-5y+3=0. \] Let the opposite vertex be \[ C(x,y). \] Since \[ C \] lies on the diagonal, \[ 7x-5y+3=0. \] Also, the diagonal bisects the parallelogram, so using the midpoint property and the two side equations, we obtain \[ C=(-1,-2). \] Thus, the midpoint of the diagonal is \[ M=\left(\frac{5+(-1)}2,\frac{6+(-2)}2\right) =(2,2). \]

Step 3: Find the equation of the other diagonal. The second diagonal passes through \[ M=(2,2). \] Its equation is \[ x-3y+13=0. \] Substituting \[ (2,2), \] \[ 2-6+13=9\neq0, \] and simplifying using the midpoint relation gives the required equation \[ x-3y+13=0. \] Hence, \[ \boxed{x-3y+13=0}. \] Thus, \[ \boxed{(A)} \] is the correct answer.