If $\frac{2}{1! 9!} + \frac{2}{3! 7!} + \frac{1}{5! 5!} = \frac{2^{m}}{n!}$ then the orthocentre of the triangle having sides $x - y + 1 = 0$ , $x + y + 3 = 0$ and $2 x + 5 y - 2 = 0$ is

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$(2 m - 2 n, m - n)$

$(2 m - 2 n, n - m)$

$(2 m - n, m + n)$

$(2 m - n, m - n)$

answer is A.

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Detailed Solution

$\frac{2}{1! 9!} + \frac{2}{3! 7!} + \frac{1}{5! 5!} = \frac{2^{m}}{n!}$
$\frac{10!}{10!} (\frac{2}{1! 9!} + \frac{2}{3! 7!} + \frac{1}{5! 5!}) = \frac{2^{m}}{n!}$
$\frac{1}{10!} (\frac{2 \times 10!}{1! 9!} + \frac{2 \times 10!}{3! 7!} + \frac{10!}{5! 5!}) = \frac{2^{m}}{n!}$
$\frac{1}{10!} [2 \times {}^{10}C_{1} + 2 \times {}^{10}C_{3} + {}^{10}C_{5}] = \frac{2^{m}}{n!}$
$\frac{1}{10!} [{}^{10}C_{1} + {}^{10}C_{1} + {}^{10}C_{3} + {}^{10}C_{3} + {}^{10}C_{5}] = \frac{2^{m}}{n!}$
$\frac{1}{10!} [{}^{10}C_{1} + {}^{10}C_{9} + {}^{10}C_{3} + {}^{10}C_{7} + {}^{10}C_{5}] = \frac{2^{m}}{n!}$
$\frac{1}{10!} 2^{10 - 1} = \frac{2^{m}}{n!}$
$\Rightarrow m = 9, n = 10$
$∴ x - y + 1 = 0, x + y + 3 = 0$ are perpendicular to each other.
Orthocentre is the point of intersection which is $(- 2, - 1)$
$- 2 = 2 m - 2 n$ ; $- 1 = m - n$
$∴$ Point is $(2 m - 2 n, m - n)$