The nearest point on the line $3 x+4 y=12$ from the origin is
The nearest point on the line $3 x+4 y=12$ from the origin is
$\left(\frac{36}{25}, \frac{48}{25}\right)$
$\left(3, \frac{3}{4}\right)$
$\left(2, \frac{3}{2}\right)$
None of these
Solution
If ' $\mathrm{D}$ ' be the foot of altitude, drawn from origin to the given line, then ${ }^{4} \mathrm{D}^{\prime}$ is the required point. Let $\angle \mathrm{OBA}=\theta$
$\Rightarrow \tan \theta=4 / 3$
$\Rightarrow \angle \mathrm{DOA}=\theta$
$\begin{aligned}
&\text { we have } \quad \mathrm{OD}=12 / 5 \text { . }\\
&\text { If } D \text { is }(h, k) \text { then } h=O D \cos \theta, k=O D \sin \theta\\
&\Rightarrow \mathrm{h}=36 / 25, \mathrm{k}=48 / 25 .
\end{aligned}$
