Let two straight lines drawn from the origin [O] intersect the line 3 x+4 y=12 at the poin…

Question

Let two straight lines drawn from the origin $$\mathrm{O}$$ intersect the line $3 x+4 y=12$ at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ such that $$\triangle \mathrm{OPQ}$$ is an isosceles triangle and $$\angle \mathrm{POQ}=90^{\circ}$$. If $$l=\mathrm{OP}^2+\mathrm{PQ}^2+\mathrm{QO}^2$$, then the greatest integer less than or equal to $l$ is :

Accepted Answer

$O P=O Q$ ($$\triangle P Q R$$ 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 $$P\left(\frac{84}{25}, \frac{12}{25}\right)$$

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

Solutions · 1