Question

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

Answer 1

Let $AB$ and $AD$ be consecutive sides of parallelogram $ABCD$ having equations $4x+5y=0$ and $7x+2y=0$ respectively. These $2$ lines intersect at $A(0,0)$

Similarly we get the co-ordinates of $B(\frac{5}{3},-\frac{4}{3})$ and $D(-\frac{2}{3},\frac{7}{3})$.

Since diagonals of a parallelogram bisect each other.

$\therefore P$ is the mid-point of $BD$.

Thus the coordinates of $P$ are-

$(\frac{1}{2},\frac{1}{2})$

Therefore, equation of $AC$ passing through point $P$ is-

$(y-0)=\frac{(\frac{1}{2}-0)}{(\frac{1}{2}-0)}(x-0)$

$\Rightarrow x=y$

Now as $BC\parallel AD$

$\therefore$ Equation of line $BC$-

$7(x-\frac{5}{3})+2(y+\frac{4}{3})=0$

$\Rightarrow 7x+2y=9$

Similarly, as $CD\parallel AB$

$\therefore$ Equation of line $CD$-

$4(x+\frac{2}{3})+5(y-\frac{7}{3})=0$

$\Rightarrow 4x+5y=9$