First let us find the midpoint of the base. As base is on #x-2y=6#, perpendicular from vertex #(1,5)# will have equation #2x+y=k# and as it passes through #(1,5)#, #k=2*1+5=7#. Hence equation of perpendicular from vertex to base is #2x+y=7#.
Intersection of #x-2y=6# and #2x+y=7# will give us midpoint of base. For this, solving these equations (by putting value of #x=2y+6# in second equation #2x+y=7#) gives us
#2(2y+6)+y=7#
or #4y+12+y=7#
or #5y=-5#.
Hence, #y=-1# and putting this in #x=2y+6#, we get #x=4#, i.e. mid point of base is #(4,-1)#.
Now, equation of a line having a slope of #3# is #y=3x+c# and as it passes through #(1,5)#, #c=y-3x=5-1*3=2# i.e. equation of line is #y=3x+2#
Intersection of #x-2y=6# and #y=3x+2#, should there give us one of the vertices. Solving them, we get #y=3(2y+6)+2# or #y=6y+20# or #y=-4#. Then #x=2*(-4)+6=-2# and hence one vertex is at #(-2,-4)#.
We know that one of the vertices on base is #(-2,-4)#, let other vertex be #(a,b)# and hence midpoint will be given by #((a-2)/2,(b-4)/2)#. But we have midpoint as #(4,-1)#.
Hence #(a-2)/2=4# and #(b-4)/2=-1# or #a=10# and #b=2#.
Hence two vertices are #(-2,-4)# and #(10,2)#
