Question

Two adjacent sides of a parallelogram are $4x+5y=0$ and $7x+2y=0$ one of its diagonals is $11x+7y-9=0$

Equation of the other diagonals is

• A. $x+y=0$
• B. $7x-11y=0$
• C. $x-y=0$
• D. $11x+7y=0$

Answer 1

Answer: (C)

$x-y=0$

Intersection of the sides will give us the included vertex while the intersection of the diagonals and the sides will give us other two vertices.
Now
$4x+5y=0$
$11x+7y-9=0$
$\to A=(\frac{5}{3},\frac{-4}{3})$
$4x+5y=0$
$7x+2y=0$
$\to B=(0,0)$
$7x+2y=0$
$11x+7y-9=0$
$\to C=(\frac{-2}{3},\frac{7}{3})$
Since diagonals bisect each other in a parallelogram
Hence it will pass through the midpoint of the of the given diagonal.
let P be the midpoint of AC
$P=\frac{1}{2},\frac{1}{2}$
Hence the other diagonal passes through P and B.
Therefore
$\frac{y-\frac{1}{2}}{x-\frac{1}{2}}=\frac{\frac{1}{2}-0}{\frac{1}{2}-0}$
$y-\frac{1}{2}=x-\frac{1}{2}$
$x-y=0$
Thus the equation of the other diagonal is $x-y=0$.