to locate a point P on AB such that AP=5/3 AB line segment AB should be divided in the ratio :
A)3:5 B)5:3 explain the answer.... plsss fast
Since AP = 5/3 AB (fraction is greater than 1) ..... This means that point P lies on the extended line AB(or say point P divides the line externally)
so we have to find the ratio in which point P divides AB
A...........B............P
Now AP = 5/3 AB
⇒ AB+BP = 5/3 AB
⇒BP = 5/3 AB - AB
⇒BP = 2/3 AB
Now AP = 5/3 AB and BP=2/3 AB
So ratio = AP/BP =(5/3)/(2/3)= (5:2) (note:- here BP =PB as we are interested in length only)
Let us illustrate this with an example
suppose Line AB is on the x-axis such that A has the cordinates(0,0), B has (3,0) and P has (5,0) satisfying the conditions given in the question
so here AP/BP=5/2
so we have to find the ratio in which point P divides AB
A...........B............P
Now AP = 5/3 AB
⇒ AB+BP = 5/3 AB
⇒BP = 5/3 AB - AB
⇒BP = 2/3 AB
Now AP = 5/3 AB and BP=2/3 AB
So ratio = AP/BP =(5/3)/(2/3)= (5:2) (note:- here BP =PB as we are interested in length only)
Let us illustrate this with an example
suppose Line AB is on the x-axis such that A has the cordinates(0,0), B has (3,0) and P has (5,0) satisfying the conditions given in the question
so here AP/BP=5/2