Three vertices of a triangle are A(4, 3),B(1, | vedclass

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(D) The centroid,orthocentre,incentre,and circumcentre of a triangle are collinear if the triangle is isosceles.For the triangle to be isosceles,the s

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$B$ or $C$ both

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(D) The centroid,orthocentre,incentre,and circumcentre of a triangle are collinear if the triangle is isosceles.For the triangle to be isosceles,the sides must satisfy $AB = BC$,$AB = AC$,or $BC = AC$.Case $1$: $AB = BC$$AB^2 = (4-1)^2 + (3-(-1))^2 = 3^2 + 4^2 = 9 + 16 = 25 \implies AB = 5$.$BC^2 = (7-1)^2 + (k-(-1))^2 = 6^2 + (k+1)^2 = 36 + (k+1)^2$.$36 + (k+1)^2 = 25 \implies (k+1)^2 = -11$ (No real solution).Case $2$: $AB = AC$$AC^2 = (7-4)^2 + (k-3)^2 = 3^2 + (k-3)^2 = 9 + (k-3)^2$.$9 + (k-3)^2 = 25 \implies (k-3)^2 = 16 \implies k-3 = \pm 4$.$k = 7$ or $k = -1$.Case $3$: $BC = AC$$36 + (k+1)^2 = 9 + (k-3)^2$.$36 + k^2 + 2k + 1 = 9 + k^2 - 6k + 9$.$37 + 2k = 18 - 6k \implies 8k = -19 \implies k = -19/8$.Since the triangle is isosceles for $k = 7, -1, -19/8$,the points are collinear in all these cases.

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

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 $\Delta ABC$ lie on the same straight line is/are:

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