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

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Solution

Q.
Let the equations of two adjacent sides of a parallelogram ABCD be $2 x - 3 y = - 23$ and $5 x + 4 y = 23$ . If the equation of its one diagonal AC is $3 x + 7 y = 23$ and the distance of A from the other diagonal is $d$ , then $50 d^{2}$ is equal to ________ . [2023]

Ans.
(529)

$We have, 2 x - 3 y = - 23$    $\dots (i)$

   $5 x + 4 y = 23$    $\dots (i i)$

and    $3 x + 7 y = 23$ $\dots (i i i)$

$Solve eqn. (i) and (iii), we get A \equiv (- 4, 5)$

$Solve eqn. (ii) and (iii), we get C \equiv (3, 2)$

$Solve eqn. (i) and (ii), we get B \equiv (- 1, 7)$

$Mid point of A C will be (- \frac{1}{2}, \frac{7}{2}) .$

$Equation of diagonal B D is$

$y - \frac{7}{2} = \frac{7 / 2}{- 1 / 2} (x + \frac{1}{2})$

$\Rightarrow 7 x + y = 0$

$Distance of A from diagonal B D is$

$d = \frac{| 7 \times (- 4) + 5 |}{\sqrt{49 + 1}} = \frac{23}{\sqrt{50}}$

$\Rightarrow 50 d^{2} = {(23)}^{2} = 529$

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