Question

The vertices of a triangle are $A(a,0),B(0,b)$ and $C(a,b)$

Answer 1

The given vertices represent a right angled triangle with right angle at $C(a,b)$ and the hypotenuse $AB$, as ${AB}^2={AC}^2+{BC}^2$
So that orthocentre of the triangle is at $C(a,b)$.
Circumcentre of the triangle is the middle point of the hypotenuse $AB$ i.e. $(a/2,b/2)$
And centroid of the triangle is
$(\frac{a+a+0}{3},\frac{0+b+b}{3})=(\frac{2a}{3},\frac{2b}{3})$
Let $P(x,y)$ be the foot of the altitude from C.
Then P lies on AB whose equation is $\frac{x}{a}+\frac{y}{b}=1$ ( 1 )
Also $CP$ is perpendicular to AB and its equation
$\Rightarrow y-b=\left(a/b\right)(x-a)$ ( 2 )
Solving (1) & (2) we get
$x=\frac{a^3}{a^2+b^2},y=\frac{b^3}{a^2+b^2}$