Question

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

Answer 1

nothing is given abt the slope

Answer 2

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)