Straight Line Question 227

Question: Let $ (h,k) $ be a fixed point where $ h>0,k>0. $ A straight line passing through this point cuts the positive direction of the coordinate axes at the points P and Q. Then the minimum area of the $ \Delta OPQ.O $ O being the origin, is

Options:

A) 4hk sq. units

B) 2hk sq. units

C) 3hk sq. units

D) None of these

Show Answer

Answer:

Correct Answer: B

Solution:

  • [b] Let the equation of any line passing through $ A(h,k) $ be $ y-k=m(x-h). $ Let this line cut the x-axis and y-axis at P and Q. Then $ P\equiv ( h-\frac{k}{m},0 ) $ and $ Q\equiv (0,k-mh). $ Let S be the area of $ \Delta OPQ, $ then $ S=\frac{1}{2}OP\times OQ=\frac{1}{2}( h-\frac{k}{m} )(k-mh) $

$ =\frac{1}{2}\frac{(mh-k)(k-mh)}{m} $

$ \Rightarrow 2mS=hkm-k^{2}-h^{2}m^{2}+khm $

$ \Rightarrow h^{2}m^{2}-2(hk-S)m+k^{2}=0 $ Since, m is real
$ \therefore $ its discriminant $ D\ge 0 $

$ \therefore 4{{(hk-S)}^{2}}-4h^{2}k^{2}\ge 0 $

$ \Rightarrow S-2hk\ge 0\Rightarrow S\ge 2hk $ Hence, minimum value of S is 2hk sq. units.