JEE 2024 — Mathematics PYQ

$Le{t}^{\prime }{Q}^{\prime }betheangleb/wtwogivenlines$

$\tan \theta =\left|\frac{\frac{2}{3}-\sqrt{3}}{1+\frac{2}{3}\sqrt{3}}\right|$

$\tan \theta =\left|\frac{2-3\sqrt{3}}{3+2\sqrt{3}}\right|$

$\tan \theta =\frac{3\sqrt{3}-2}{2\sqrt{3}+3}$

$Now$

$Pm=\left|\frac{2(2)-3(3)+28}{\sqrt{4+9}}\right|$

$=\frac{23}{\sqrt{13}}$

$\therefore PQ=\frac{Pm}{\sin \theta }$

$=Pm\csc \theta$

$=\frac{23}{\sqrt{13}}\left(\frac{\sqrt{(2\sqrt{3}+3{)}^2+(3\sqrt{3}-2{)}^2}}{3\sqrt{3}-2}\right)$

$=\frac{23}{\sqrt{13}}\left(\frac{\sqrt{12+9+12\sqrt{3}+27+4-12\sqrt{3}}}{3\sqrt{3}-2}\right)$

$=\frac{23}{\sqrt{13}}\times \frac{\sqrt{52}}{(3\sqrt{3}-2)}$

$=\frac{46}{3\sqrt{3}-2}\times \frac{3\sqrt{3}+2}{3\sqrt{3}+2}$

$=\frac{46(3\sqrt{3}+2)}{23}$

$=4+6\sqrt{3}$