Let #D# be the endpt. of altd. thro. vertex #C(6,3).# Then, #D# lies on the side#AB.#
Thus, #D# is the pt. of intersection of altd.#CD# with side #AB#
Accordingly, to find #D#, we have to find eqns. of #AB# and #CD# and solve them.
Eqn. of Side AB :-
With #A(8,7)# and, #B(1,5)#, slope of #AB# is #(7-5)/(8-1)=2/7#
#B(1,5)# iis on #AB#. Hence, eqn. of #AB# #:y-5=2/7(x-1),# i.e., #y=5+2/7(x-1)............(i)#
Eqn. of Altd. CD :-
#CD# is perp. to #AB#, and, slope of #AB# is #2/7#, so, the slope of #CD# has to be #-1/(2/7)=-7/2,# & with #C(6,3)# on it, we have, eqn. of #CD : y-3=-7/2(x-6),# or, #y=3-7/2(x-6).................(ii)#
To get #D#, solving #(i) & (ii) : 5+2/7(x-1)=3-7/2(x-6)rArr70+4x-4=42-49x+294rArr4x+49x=42+294+4-70rArr53x=270rArrx=270/53#
Then, by #(i), y=5+2/7(270/53-1)=5+2/7*217/53=5+62/53=327/53#
So, the end-pt. of Altd. from corner #C# is #D(270/53,327/53)#.
Finally, length of Altd. #CD# = Perp. Dist. from pt. #C(6,3)# to side #AB : y=5+2/7(x-1)#
#=|3-5-2/7(6-1)|/sqrt(1+4/49)=|2+10/7|/sqrt(53/49)=24/sqrt53.#
