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

Question

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

Accepted Answer

$OP=OQ$ ($	rianglePQR$ is isosceles triangle)

Let slope of line $OP	o {m}_{1}$

So, equation $	o y={m}_{1}x$

$\begin{gathered} 	an {45}^{\circ }=|\frac{{m}_{1}-{m}_{2}}{{m}_{1}{m}_{2}}| \ 1=|\frac{{m}_{1}+\frac{3}{4}}{1-\frac{3}{4}{m}_{1}}| \ \Rightarrow 1-\frac{3}{4}{m}_{1}={m}_{1}+\frac{3}{4} \ \frac{1}{4}=\frac{7}{4}{m}_{1}\Rightarrow {m}_{1}=\frac{1}{7} \ OP	o y=\frac{1}{7}x \end{gathered}$

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

$\begin{gathered} \Rightarrow \, O{P}^{2}={a}^{2}={(\frac{84}{25})}^{2}+{(\frac{12}{25})}^{2}=\frac{288}{25} \ 	herefore \, I=O{P}^{2}+P{Q}^{2}+Q{O}^{2} \ ={a}^{2}+{a}^{2}+2{a}^{2} \ =4{a}^{2} \ =4	imes \frac{288}{25} \ I=46.08 \ [I]=46 \end{gathered}$

Solutions · 1