Question

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

Answer 1

The sides are $4x+5y=0$ ….$\left(1\right)$

and $7x+2y=0$ …$\left(2\right)$

Equation of the diagonal,

$11x+7y=9$ …$\left(3\right)$

Solving $\left(1\right)$ and $\left(3\right)$

$4x+5y=0\Rightarrow 4x=-5y$ or $x=\frac{-5y}{4}$

$11x+7y=9\Rightarrow 11\times \frac{-5y}{4}+7y=9$

$\Rightarrow \frac{-55y+28y}{4}=9$

$\Rightarrow -27y=36$

$\Rightarrow y=\frac{-36}{27}=\frac{-4}{3}$

Substituting $y=\frac{-4}{3}$ in $4x+5y=0$ we get

$4x=-5y$ or $4x=-5\times \frac{-4}{3}=-\frac{-20}{3}$

$\Rightarrow x=\frac{5}{3}$

co-ordinates of $A$ are $(\frac{5}{3},\frac{-4}{3})$

Solving $\left(2\right)$ and $\left(3\right)$

$7x+2y=0\Rightarrow x=-\frac{2y}{7}$

$11x+7y=9\Rightarrow 11\times \frac{-2y}{7}+7y=9$

$\Rightarrow \frac{-22y}{7}+7y=9$

$\Rightarrow -22y+49y=63$

$\Rightarrow 27y=63$

$\Rightarrow y=\frac{63}{27}=\frac{7}{3}$

Substituting $y=\frac{7}{3}$ in $7x+2y=0$ we get

$7x=-2y=-2\times \frac{7}{3}$

$\Rightarrow x=\frac{-2}{3}$

co-ordinates of $B$ are $(\frac{-2}{3},\frac{7}{3})$

Mid-point of $AB$ is $M(\frac{\frac{5}{3}+\frac{-2}{3}}{2},\frac{\frac{-4}{3}+\frac{7}{3}}{2})=(\frac{1}{2},\frac{1}{2})$

$\therefore$ equation of the diagonal $OC$ is $y=x$