The nearest point on the line $3 x+4 y=12$ from the origin is

• Correct answer
  A

  $\left(\frac{36}{25}, \frac{48}{25}\right)$
• B

  $\left(3, \frac{3}{4}\right)$
• C

  $\left(2, \frac{3}{2}\right)$
• D

  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 $\angle \mathrm{OBA}=\theta$ $\Rightarrow \tan \theta=4 / 3$ $\Rightarrow \angle \mathrm{DOA}=\theta$ $\begin{aligned} &\text { we have } \quad \mathrm{OD}=12 / 5 \text { . }\\ &\text { If } D \text { is }(h, k) \text { then } h=O D \cos \theta, k=O D \sin \theta\\ &\Rightarrow \mathrm{h}=36 / 25, \mathrm{k}=48 / 25 . \end{aligned}$

Hence, the correct answer is option: 'A'.