Let #D# be the end.pt. of altd. from vertex #C(2,4)# on side #AB# of #Delta ABC#.
Thus, pt.#D# is the pt. of intersection of altd. #CD# with side #AB.#
Accordingly, to find co-ords. of #D#, we have to solve the eqns. of #CD# & #AB.#
Eqn. of #AB#:-
Given that #A(4,5), B(3,6),# so, eqn. of #AB# is given by,
#(y-6)/(5-6)=(x-3)/(4-3),# i.e., #y-6+x-3=0,# or,
#x+y-9=0......(1)#
Eqn. of #CD#:-
Recall that eqn. of a line, perp. to line #ax+by+c=0, # is #bx-ay+k=0# for some #k in RR#
Altd. #CD# is perp. to #AB# having eqn. (1), so, its eqn. can be taken as #y-x+k=0.# To determine #k in RR,# we use the fact that #C(2,4) in CD,# so, we must have #4-2+k=0,# giving #k=-2#, and thus, #CD# is given by #y-x-2=0.................(2)#
Solving #(1) & (2)#, we get #D(7/2,11/2)#
The reqd. end-pts. of altd. #CD# are #C(2.4) & D(7/2,11/2).#
Length of altd. #CD#= the dist. btwn. #C &D
=sqrt{(7/2-2)^2+(11/2-4)^2}=sqrt{9/4+9/4)=3/2sqrt2.#
Alternatively , dist. #CD# can be obtained by using the following formula :-
Perp. dist. btwn. a pt. #(h,k)# & line #ax+by+c=0,# is #|ah+bk+c|/sqrt(a^2+b^2)#
Here, #(h,k)=(2,4)# & line #AB# : #ax+by+c+0# is #x+y-9=0.#
#:.# Dist. #CD=|2+4-9|/(sqrt(1^2+1^2))=3/sqrt2=3/2sqrt2,# as before!
