Question

A straight line L with negative slope passes through
the point (9,4) and cuts the positive coordinate axes
at the points P and Q respectively,
Area of AOPQ, when OP + O2 becomes
minimum is (where O is the origin)
75 units
225 units
125 units
200 untis​

Answer 1

Given :  A straight line L with negative slope passes through  the point (9,4) and cuts the positive coordinate axes  at the points P and Q  respectively,

To find : Area of ΔOPQ, when OP + OQ becomes  minimum

Solution:

Area of Triangle  Δ OPQ  = (1/2) OP * QQ

Let say Line =  y = -mx  + c

passes though 9 , 4

=> 4 = -9m + c

=> c = 9m + 4

y = -mx  + 9m + 4

P = 0 ,  9m + 4

Q = (9m + 4)/ m  , 0

OP = 9m + 4

OQ =  (9m + 4)/ m

OP + OQ  =  9m + 4 + (9m + 4)/ m

L  = 9m + 13   + 4/m

dL/dm = 9  - 4/m²

9  - 4/m² == 0

=> m = 2/3   ( m is + ve as -ve sign for slope already taken)

d²L/dm²  =  8/m³ > 0

=> L is minimum

Hence OP + OQ is minimum when  m =  2/3

OP = 9(2/3) + 4  =  10

OQ = 10/(2/3)  = 15

Area = (1/2) 10 * 15  =  75  sq units

Area of ΔOPQ, when OP + OQ becomes  minimum is 75  sq units

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