JEE Main 2024 — 5 April, Shift 1 — Question 3

$\text{Given\; line:\; } 3x + 4y = 12,\; \text{ points\; } P, Q\; \text{ lie\; on\; it,\; lines\; from\; origin\; have\; slopes\; } m_1, m_2.$

$\text{Coordinates\; of \; intersection: }$ $(x, y) = \left( \frac{12}{3 + 4 m}, \frac{12 m}{3 + 4 m} \right)$

$\text{Condition\; for\; right\; angle\; at\; origin:\; } m_1 m_2 = -1$

$\text{Condition\; for \; isosceles\; triangle:\; } |OP| = |OQ| \implies \frac{1+m_1^2}{(3+4 m_1)^2} = \frac{1+m_2^2}{(3+4 m_2)^2}, \ m_2 = -\frac{1}{m_1}$

$\text{Solve:\; } (3+4 m_1)^2 = (3 m_1 - 4)^2 \implies m_1 = -7 \; \text{ or \;} m_1 = \frac{1}{7} \implies m_2 = \frac{1}{7}\; \text{ or\; } m_2 = -7$

$\text{Coordinates: }$ $P = \left(-\frac{12}{25}, \frac{84}{25}\right), \quad$ $Q = \left(\frac{84}{25}, \frac{12}{25}\right)$

$\text{Compute squared lengths:}$

OP^2 &= \left(-\frac{12}{25}\right)^2 + \left(\frac{84}{25}\right)^2 = \frac{7200}{625} = 11.52 \\ OQ^2 &= \left(\frac{84}{25}\right)^2 + \left(\frac{12}{25}\right)^2 = \frac{7200}{625} = 11.52 \\ PQ^2 &= \left(\frac{84+12}{25}\right)^2 + \left(\frac{12-84}{25}\right)^2 = \frac{14400}{625} = 23.04 \end{aligned}$$ $\text{Sum:\; } l = OP^2 + OQ^2 + PQ^2 = 11.52 + 11.52 + 23.04 = 46.08$ $\text{Greatest\; integer \;less\; than\; or\; equal\; to\; } l:$

\boxed{46}

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