The base of a triangle is axis of x and its other 2 sides are given by the equations :y=[(l+α)/α]x + (1+α) and y=[(l+ß)/ß]x+(1+ß).Prove that the locus of its orthocenter is the line x+y=0

The base of a triangle is axis of x and its other 2 sides are given by the equations :y=[(l+α)/α]x + (1+α) and y=[(l+ß)/ß]x+(1+ß).Prove that the locus of its orthocenter is the line x+y=0

Solution :Steps 

Step 1:Follow this figure 

Step 2:

Step 3:

The eqaution of line which is perpendicular to a given line ax+by+c=0 is bx-ay+k=0 .This deduction would clarify the value of (x,y)=(αβ,-αβ) is the point where perpendicular from C to AB intersect line II.