Mathematics - Straight Lines Question with Solution | TestHub

$\begin{array}{rl}& 3\mathrm{x}+4\mathrm{y}=12\\ & 3(\mathrm{r}\cos \theta )+4(\mathrm{r}\sin \theta )=12\\ & \mathrm{r}(3\cos \theta +4\sin \theta )=12\dots (1)\\ & 3(-\mathrm{r}\sin \theta )+4(\mathrm{r}\cos \theta )=12\\ & \mathrm{r}(-3\sin \theta +4\cos \theta )=12\dots (2)\\ & {\left(\frac{12}{\mathrm{r}}\right)}^2+{\left(\frac{12}{\mathrm{r}}\right)}^2=(3\cos \theta +4\sin \theta {)}^2+(-3\sin \theta +4\cos \theta {)}^2\\ & 2{\left(\frac{12}{\mathrm{r}}\right)}^2=9+16\\ & \frac{2\times 144}{{\mathrm{r}}^2}=25\Rightarrow 288=25{\mathrm{r}}^2\\ & \Rightarrow \frac{288}{25}={\mathrm{r}}^2\\ & \Rightarrow \sqrt{2}\left(\frac{12}{5}\right)=\mathrm{r}\\ & \ell ={\mathrm{OP}}^2+{\mathrm{PQ}}^2+{\mathrm{QO}}^2\\ & \ell ={\mathrm{r}}^2+{\mathrm{r}}^2+{\mathrm{r}}^2(\cos \theta +\sin \theta {)}^2+{\mathrm{r}}^2(\sin \theta +\cos \theta {)}^2\\ & =2{\mathrm{r}}^2+{\mathrm{r}}^2(1+\sin 2\theta +1-2\sin 2\theta )\\ & =2{\mathrm{r}}^2+2{\mathrm{r}}^2\\ & =4{\mathrm{r}}^2\\ & =4\left(\frac{288}{25}\right)=\frac{1152}{25}=46.08\\ & [\ell ]=46\end{array}$