a straight line with negative slope passes through the point (9,4) and cuts the positive co ordinate at P and Q respectively
i) min value of OP+OQ (where O is origin)
ii)Area of OPQ when OP+OQ is min

Question

anonymous · Answer

nothing is given abt the slope

anonymous · Answer

Take slope to be -m then eq of line will be

$$mx+y-4-9m=0$$

You get P(9m+4/m,0) and Q(0,9m+4)
so you get, $$OP + OQ = (9m+4)(1+1/m)$$

differentiating with resp to m you will get $$d(OP+OQ)/dm=9-4/m^2$$

and on equating it with zero you will get m= +2/3,-2/3
take the negative slope and find OP and OQ

for the second part you need to use Ar(triangle)=1/2 *(OP)*(OQ)