QJEEM24Ma8236: JEE Main 2024 Mathematics - Class 11 JEEM24Ma8236 Solution: Let two straight lines drawn from the origin intersect the line …

$O P=O Q$ ( $\triangle P Q R$ is isosceles triangle)

Let slope of line $O P \rightarrow m_1$

So, equation $\rightarrow y=m_1 x$

$\begin{aligned}&\begin{aligned}& \tan 45^{\circ}=\left|\frac{m_1-m_2}{m_1 m_2}\right| \\& 1=\left|\frac{m_1+\frac{3}{4}}{1-\frac{3}{4} m_1}\right| \\& \Rightarrow 1-\frac{3}{4} m_1=m_1+\frac{3}{4} \\& \frac{1}{4}=\frac{7}{4} m_1 \Rightarrow m_1=\frac{1}{7}\end{aligned}\\&\text { Equation } O P \rightarrow y=\frac{1}{7} x\end{aligned}$

Point of intersection of $O P$ & line $3 x+4 y=12$ is $P\left(\frac{84}{25}, \frac{12}{25}\right)$

$\begin{aligned}& \Rightarrow \quad O P^2=a^2=\left(\frac{84}{25}\right)^2+\left(\frac{12}{25}\right)^2=\frac{288}{25} \\& \therefore \quad I=O P^2+P Q^2+Q O^2 \\&=a^2+a^2+2 a^2 \\&=4 a^2 \\&=4 \times \frac{288}{25} \\& I=46.08 \\& {[I] }=46\end{aligned}$