Question

In a triangle formed by the lines $xy=0$ and $2x+3y=k$

Answer 1

Since, the triangle is formed by the lines $xy=0$ and $\frac{x}{\left(k/2\right)}+\frac{y}{\left(k/3\right)}=1$
and pair of straight lines $xy=0$ are perpendicular to each other
Therefore, origin is the orthocenter of the triangle for all values of $k$
Vertices of triangle are $A(0,0)$, $B(\frac{k}{2},0)$ and $C(0,\frac{k}{3})$
Therefore, centroid $G=(3,2)=(\frac{k}{6},\frac{k}{9})$
$\Rightarrow k=18$
Since, $\Delta BAC$ is an right-angled
Therefore, circumcenter(C) is the midpoint of side $BC$
$C=(3,2)=(\frac{k}{4},\frac{k}{6})$
$\Rightarrow k=12$
Area of $\Delta ABC=192=\frac{k^2}{12}$
$\Rightarrow k=\pm 48$