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. $x+2y=0$
• B. $2x+y=0$
• C. $x-y=0$
• D. $Noneofthese$

Answer 1

The correct option is C $x-y=0$

Given two consecutive sides of a parallelogram are $4x+5y=0,7x+2y=0$

The point of intersection of these sides is $(0,0)$

The equation of one diagonal is $11x+7y=9$

The point of intersection of $4x+5y=0$ and $11x+7y=9$will be $11(-\frac{5y}{4})+7y=9⟹y=-\frac{4}{3}⟹x=\frac{5}{3}$

The point of intersection of $7x+2y=0$ and $11x+7y=9$will be $11(-\frac{2y}{7})+7y=9⟹y=\frac{7}{3}⟹x=-\frac{2}{3}$

The mid point of $(\frac{5}{3},-\frac{4}{3})$ and $(-\frac{2}{3},\frac{7}{3})$ is $(\frac{1}{2},\frac{1}{2})$

Let the fourth vertex be $(h,k)$

$(\frac{1}{2},\frac{1}{2})$ is the mid point of $(0,0)$ and $(h,k)$

$⟹\frac{h}{2}=\frac{1}{2}⟹h=1,\frac{k}{2}=\frac{1}{2}⟹k=1$

So the other diagnol is $\frac{y-0}{x-0}=\frac{1-0}{1-0}⟹x-y=0$