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−22,b−42). But we have midpoint as (4,−1).
Hence a−22=4 and b−42=−1 or a=10 and b=2.
Hence two vertices are (−2,−4) and (10,2)
