Refer to the figure attached.
The point (3,√3) does not satisfy the equation of the altitude √3x+y=4.
Hence it lies on the base of the equilateral triangle to which the altitude is drawn. Let B be the point (3,√3) and AD be the altitude.
Therefore BC is perpendicular to the altitude. Hence equation of BC is given by,
(y−√3)=(x−3)√3
x=√3y
Substituting int the equation of the altitude we get the mid point of BC as (32,√32) ...(i)
Therefore If B=(3,√3) and the mid point of BC is (32,√32) ...(from i)
Therefore C is (0,0).
Now we know that the side AB and AC make angles of 300 with the altitude.
Therefore considering the slope of AB or AC as m and the slope of the altitude being −√3 we get,
tan(300)=∣∣∣m+√31−√3m∣∣∣
⇒±1√3=m+√31−√3m
⇒m=−1√3orm=∞
Taking m=−1√3 as slope of AB equation of side AB is,
(y−√3)=−x+3√3
√3y−3=−x+3
x+√3y=6 ...(ii)
Solving the eqautions of the altitude of the triangle and the side AB, we get A as,
A=(0,2√3)
Therefore summing up we get,
A=(0,2√3) ; B=(3,√3) ; C=(0,0)
Similarly, taking m=∞ as slope of AB we get equation of AB as x=3,
Again solving equation of AB and AD, we get coordinates of A in this case as (3,−√3)
For this case the coordinates are,
A:(3,−3√3) ; B:(3,√3) ; C:(0,0)
The two possible triangles are shown in the figure.
For an equilateral triangle, orthocenter and centroid coincide.
For the first triangle centroid≡Orthocenter≡(33,3√33)
Hence, (1,√3) can be the orthocenter.