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  • Let the equations of two adjacent sides of a parallelogram <img alt="\mathrm{ABCD}$ be $2 x-3 y=-23" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BABCD%7D%24%20be%20%242%20x-3%20

Let the equations of two adjacent sides of a parallelogram \mathrm{ABCD}$ be $2 x-3 y=-23 and 5 x+4 y=23. If the equation of its one diagonal AC is3 \mathrm{x}+7 \mathrm{y}=23 and the distance of A from the other diagonal is \mathrm{d}, then 50 \mathrm{~d}^2 is equal to _________.

Option: 1

529


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

A & C point will be (–4, 5) & (3, 2 )

mid point of AC will be \left(-\frac{1}{2}, \frac{7}{2}\right)

equation of diagonal BD is

\begin{aligned} & y-\frac{7}{2}=\frac{\frac{7}{2}}{-\frac{1}{2}} \quad\left(\mathrm{x}+\frac{1}{2}\right) \\ & \Rightarrow 7 x+y=0 \end{aligned}

Distance of A from diagonal BD

\begin{aligned} & =d=\frac{23}{\sqrt{50}} \\ & \Rightarrow \quad 50 d^2=(23)^2 \\ & 50 d^2=529 \end{aligned}

Posted by

Sanket Gandhi

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