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Q.
Two consecutive sides of a parallelogram are $4 x+5 y= 0$ and $7 x+2 y=0$. If the equation to one diagonal is $11 x + 7 y=9$, then the equation of the other diagonal is
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Solution:
Let the equation of sides $A B$ and $A D$ of the parallelogram $A B C D$ be
$4x + 5y = 0$ .....(1)
and $7x + 2y = 0$ .... (2)
respectively. Solving (1) and (2), we have $x = 0, y = 0 $
$\therefore A \equiv (0, 0)$
Equation of one diagonal of the parallelogram is
$11 x+7 y=9 $ ...... (3)
Clearly, $A(0,0)$ does not lie on diagonal (3), therefore (3) is the equation of diagonal $B D$.
Solving (1) and (3), we get
$B \equiv\left(\frac{5}{3},-\frac{4}{3}\right)$ and $D \equiv\left(-\frac{2}{3}, \frac{7}{3}\right)$.
Since $H$ is the middle point of $B D $
$\therefore H \equiv\left(\frac{1}{2}, \frac{1}{2}\right)$.
Now the equation of diagonal $A C$ which passes through
$A(0,0) $ and $ H\left(\frac{1}{2}, \frac{1}{2}\right) $ is
$ y-0=\frac{0-\frac{1}{2}}{0-\frac{1}{2}}(x-0) $
or $y-x=0$
