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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 points P and Q.find the minimum area of the triangle OPQ`
7 years ago

22 Points
```										Hi Rahul,
lets take the equation of line to be x/a + y/b = 1 where line intersects x axia at ( a, 0 ) and y axis at (0,b)
Since line passes thru (h,k) therefore h/a + k/b = 1. ------- (1)
A = area of triangle OPQ = 1/2 * a*b
from eqn (1)
A = 1/2 * b * hb / ( b- K ) ( putting value of a from (1) into A )
= 1/2*b2h/(b - k)
to calculate min value of A, dA/db = 1/2 * h * ( b2 - 2bk ) / ( b-k )2
dA / db = 0 when b2 - 2bk = 0
or b -2k = 0
or b = 2k
At this value of b = 2k , the d2A / db2 is positive that is Minimum of A occurs at this value.
Putting in (1) a = 2h
Thus Minimum value of A occurs when a = 2h and b = 2k
Hence Min(A) = 1/2 * a * b = 1/2 * 2h * 2k = 2hk
Thus Minimum area of traingle OPQ is 2hk.
Puneet

```
7 years ago
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