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# Let (h,k) be a fixed point where h,k>0. a straight line passing through this point cuts the positive direction of coordinate axes at point P and Q. Find the minimum area of the triangle OPQ, O being the origin.I FOUND THE AREA TO BE=1/2*[2hk+(h)squared*tan(theta)+(k)squared*cot(theta)]but the answer is 2hk..... how the area can be minimised???

28 Points
11 years ago

Dear Parth Tiwari:

Ans:- Let the slope is m. then the equation of the line is y-k=m(x-h) It cuts the co ordinate axix at the points

A={(k-mh), 0} and B={0,(h-m/k)} hence the area is Δ=1/2(k-mh)(h-m/k) then applying min condition for the variable m we get the min area is =2hk.

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