Let the equations of two adjacent sides of a parallelogram ABCD be 2x-3y=-23 and 5x+4y=23.

Question

Let the equations of two adjacent sides of a parallelogram $ABCD$ be $2x-3y=-23$ and $5x+4y=23$. If the equation of its one diagonal $AC$ is $3x+7y=23$ and the distance of A from the other diagonal is $d$, then $50{d}^{2}$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 529

Accepted Answer

We have, $ABCD$ is a parallelogram

Let equation of $AB$ be $2x-3y=-23\ldots$ (i)

and equation of $BC$ be $5x+4y=23\ldots$ (ii)

Equation of $AC$ is $3x+7y=23\ldots$ (iii)

Solving Eqs. (i) and (ii), we get

$x=-1$, and $y=7$

$	herefore $ Co-ordinate of $B$ is $(-1,7)$

On solving Eqs. (ii) and (iii), we get

$x=3,y=2$

$	herefore $ Co-ordinate of $C$ is $(3,2)$

On solving Eqs. (i) and (iii), we get $x=-4$ and $y=5$

$	herefore $ Co-ordinate of $A$ is $(-4,5)$.

Let $E$ be the intersection point of diagonal co-ordinate of

$E$ is $(\frac{-4+3}{2},\frac{5+2}{2})$ or $(-\frac{1}{2},\frac{7}{2})$

$\because E$ is mid-point of $AC$

$\begin{gathered} BDy-7=(\frac{7-\frac{7}{2}}{-1+\frac{1}{2}})(x+1) \ \Rightarrow 7x+y=0 \end{gathered}$

Distance of $A$ from diagonal $BD=\frac{|7	imes (-4)+5|}{\sqrt{{7}^{2}+{1}^{2}}}$

$	herefore d=\frac{23}{\sqrt{50}}$

Hence, $50{d}^{2}=(23{)}^{2}=529$

Solutions · 1