A straight line L with negative slope passes thru the pt {8,2} nd cuts the positive coordinate axes at pts P & Q. find the absolute min. value of OP + OQ ,as L varies where O is the origin.
let the equation for the line which passes thru {8,2} be y=mx+c
OP = c (on Y axis , x=0); similarly OQ= -c/m OP+OQ = c-c/m As the line passes thru {8,2}, 2=8m+C i.e c=2-8m substituting this value for OP+OQ OP+OQ= 2-8m -(2-8m)/m or OP+OQ = 2 -8m - 2/m +8 differentiatinh w.rto m , and putting it equal to zero, we get m2 =1/4 or m= -1/2 (given negative slope) and c=6 line equation is y=-0.5x+6 OP+OP =18 (this is the minimum value)
Moderation Team
Community Discussion Question:
![[Post New]](/templates/default/images/icon_minipost_new.gif){kind=icon}
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
6
