Question
$ABCD$ is rectangle with $A( - 1,2),B(3,7)$ and $AB:BC = 4:3$. If $d$ is the distance of origin from the intersection point of diagonals of rectangle, then possible values of [ $d$ ] is/are (where [.] denote greatest integer function)
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Solution
Sol. [Key Concept: $d$ = half a diagonal of the rectangle; $\lbrack d\rbrack$ takes the two listed integer values.]
$$\begin{matrix} BC & \ = \frac{3}{4}AB = \frac{3}{4}\sqrt{41} \\ P & \ = \frac{1}{2}BC = \frac{3}{8}\sqrt{41} \\ tan\theta & \ = - \frac{4}{5} \end{matrix}$$
Intersection point of diagonals can be $\left( 1 + \frac{3}{8}\sqrt{41}\left( - \frac{5}{\sqrt{41}} \right),\frac{9}{2} + \frac{3}{8}\sqrt{41}\left( \frac{4}{\sqrt{41}} \right) \right)$,
or $\left( 1 - \frac{3}{8}\sqrt{41}\left( - \frac{5}{\sqrt{41}} \right),\frac{9}{2} - \frac{3}{8}\sqrt{41}\left( \frac{4}{\sqrt{41}} \right) \right)$
i.e. $\left( - \frac{7}{8},6 \right)$ or $\left( \frac{23}{8},3 \right)$
$\therefore\ d = \sqrt{\frac{49}{64} + 36}$ or $\sqrt{\frac{529}{64} + 9}$
Hence, $\lbrack d\rbrack = 6$ and 4
✘ (a) $3$ is not an attained value of $\lbrack d\rbrack$.
✔ (b) $4$ is attained.
✘ (c) $5$ is not attained.
✔ (d) $6$ is attained.
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