Let, #P=P(x,y)# be the reqd. pt.
Then, by what is given, we need to find #P,# on the sgmt. #AB,# s.t.,
#AP=3/5*AB rArr AB=5/3*AP..............(1).#
Since, the pts. #A, P, and B# are collinear, and as, #P# lies btwn.
#A, and, B,# we have, #AP+PB=AB.#
By #(1),# then, #AP+PB=5/3*AP rArr PB=2/3*AP, i.e.,#
# (AP)/(PB)=3/2.#
This means that, the pt. #P# divides the sgmt. #AB# in the ratio
#3:2# from the pt. #A.#
Recall that, the co-ords. of a pt. #P(x,y)# that divides the sgmt.
joining #A(x_1,y_1), and, B(x_2,y_2)# in the ratio #lambda:1# from the
pt. #A,# are given by,
#x=(lambdax_2+x_1)/(lambda+1), y=(lambday_2+y_1)/(lambda+1).#
Accordingly, we get,
# x=(3/2*3+2)/(3/2+1), y=(3/2*(-3)+11)/(3/2+1), # giving,
# P(x,y)=P(13/5,13/5),# as the desired pt.
Enjoy Maths.!
