Three vertices of a triangle are  $A(4,3); B(1,-1),$ and  $C(7,k).$ The value(s) of $k$ for which the centroid, orthocentre, incentre, and circumcentre of the △ABC lie on the same line is/are:

The vertices of the triangle are A(4,3), B(1,−1), C(7,k).

Since the centroid, orthocentre, incentre, and circumcentre of the △ABC lie at the same line, △ABC .is isosceles

∴ AB = AC

$\Rightarrow {\mathrm{AB}}^2={\mathrm{AC}}^2$

$\Rightarrow {\left(\sqrt{{\left(1-4\right)}^2+{\left(-1-3\right)}^2}\right)}^2={\left(\sqrt{{\left(7-4\right)}^2+{\left(\mathrm{k}-3\right)}^2}\right)}^2$  [Using distance formula]

$\Rightarrow 9+16=9+{\left(\mathrm{k}-3\right)}^2$

$\Rightarrow {\left(\mathrm{k}-3\right)}^2=16$

$\therefore \mathrm{k}-3=4 \mathrm{or} \mathrm{k}-3=-4$

$\Rightarrow \mathbf{k}\mathbf{=}\mathbf{7} \mathrm{or} \mathbf{k}\mathbf{=}\mathbf{-}\mathbf{1}$