Supoose that, #C=C(a,b)#.
Given that, #A=A(1,1), and B=B(-1,4)#.
Let us denote, by #Grad(AB)#, the gradient of #AB#.
Then, #Grad(AB)=(4-1)/(-1-1)=-3/2=-3m........."[Given]"#.
#:. m=1/2...........................................................................(1)#.
#"Given that "Grad(AC)=3m, &, (1) rArr (b-1)/(a-1)=3/2#,
#:. 2b-2=3a-3, or, 3a-2b=1...............................(2)#.
Likewise, #Grad(BC)=m, &, (1)rArr(b-4)/(a-(-1))=(b-4)/(a+1)=1/2,#
#:. 2b-8=a+1, i.e., 2b-a=9....................................(3)#.
#(2)+(3) rArr 2a=10 rArr a=5#
#a=5, and, (3) rArr b=7#.
Thus, #m=1/2, and, C=C(a,b)=C(5,7)#.