Let A(alpha, -2),B(alpha, 6),and Cleft(frac{a | vedclass

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(B) Given vertices are $A(\alpha, -2)$,$B(\alpha, 6)$,and $C\left(\frac{\alpha}{4}, -2\right)$.Since $AB$ is a vertical line $(x = \alpha)$ and $AC$ i

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perimeter is $25$

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(B) Given vertices are $A(\alpha, -2)$,$B(\alpha, 6)$,and $C\left(\frac{\alpha}{4}, -2ight)$.Since $AB$ is a vertical line $(x = \alpha)$ and $AC$ is a horizontal line $(y = -2)$,$	riangle ABC$ is a right-angled triangle at $A$.The circumcentre of a right-angled triangle is the midpoint of the hypotenuse $BC$.Midpoint of $BC = \left(\frac{\alpha + \frac{\alpha}{4}}{2}, \frac{6 - 2}{2}ight) = \left(\frac{5\alpha}{8}, 2ight)$.Given circumcentre is $\left(5, \frac{\alpha}{4}ight)$.Equating the coordinates: $\frac{5\alpha}{8} = 5 \implies \alpha = 8$ and $\frac{\alpha}{4} = 2 \implies \alpha = 8$.Thus,$A(8, -2)$,$B(8, 6)$,and $C(2, -2)$.Lengths of sides: $AB = |6 - (-2)| = 8$,$AC = |8 - 2| = 6$,and $BC = \sqrt{8^2 + 6^2} = 10$.Area $= \frac{1}{2} 	imes AB 	imes AC = \frac{1}{2} 	imes 8 	imes 6 = 24$.Perimeter $= AB + AC + BC = 8 + 6 + 10 = 24$.Circumradius $R = \frac{BC}{2} = \frac{10}{2} = 5$.Inradius $r = \frac{AB + AC - BC}{2} = \frac{8 + 6 - 10}{2} = \frac{4}{2} = 2$.Comparing with options,the perimeter is $24$,not $25$. Thus,option $B$ is incorrect.

Let $A(\alpha, -2)$,$B(\alpha, 6)$,and $C\left(\frac{\alpha}{4}, -2\right)$ be the vertices of a $\triangle ABC$. If $\left(5, \frac{\alpha}{4}\right)$ is the circumcentre of $\triangle ABC$,then which of the following is $NOT$ correct about $\triangle ABC$?

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