Question

Two consecutive sides of a parallelogram are $4x+5y=0$ and $7x+2y=0$. If the equation to one diagonal is $11x+7y=9$, then the equation of the other diagonal is

• A. $x-y=0$
• B. $x+y=0$
• C. $x+y=1$
• D. $x-y=1$

Answer 1

Answer: (A)

Let the equation of sides $AB$ and $AD$ of the parallelogram $ABCD$ 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
$11x+7y=9$ ...... (3)
Clearly, $A(0,0)$ does not lie on diagonal (3), therefore (3) is the equation of diagonal $BD$.
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 $BD$
$\therefore H\equiv \left(\frac{1}{2},\frac{1}{2}\right)$.
Now the equation of diagonal $AC$ 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 $y-x=0$