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BITSAT 2012
Exam
BITSAT
Year
2012
Subject
Mathematics
Chapter
Straight Lines

BITSAT 2012 · Mathematics · Straight Lines

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

The nearest point on the line 3x+4y=123 x+4 y=12 from the origin isThe nearest point on the line 3 x+4 y=12 from the origin is

Options

  1. A(3625,4825)\left(\frac{36}{25}, \frac{48}{25}\right)
  2. B(3,34)\left(3, \frac{3}{4}\right)
  3. C(2,32)\left(2, \frac{3}{2}\right)
  4. DNone of these

Correct answer

A. (3625,4825)\left(\frac{36}{25}, \frac{48}{25}\right)

Step-by-step solution

If ' D\mathrm{D} ' be the foot of altitude, drawn from origin to the given line, then 4D′{ }^{4} \mathrm{D}^{\prime} is the required point. Let ∠OBA=θ⇒tan⁡θ=4/3⇒∠DOA=θ\angle \mathrm{OBA}=\theta \Rightarrow \tan \theta=4 / 3 \Rightarrow \angle \mathrm{DOA}=\theta we have OD=12/5 .  If D is (h,k) then h=ODcos⁡θ,k=ODsin⁡θ⇒h=36/25,k=48/25.\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}$$ The nearest point on the line 3 x+4 y=12 from the origin is
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