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Q. Let $A (\alpha,-2), B(\alpha, 6)$ and $C \left(\frac{\alpha}{4},-2\right)$ be 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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Solution:

$A (\alpha,-2): B (\alpha, 6): C \left(\frac{\alpha}{4},-2\right)$
since $AC$ is perpendicular to $AB$.
So, $\triangle ABC$ is right angled at $A$.
Circumcentre$=\text { mid point of } BC .=\left(\frac{5 \alpha}{8}, 2\right) $
$ \therefore \frac{5 \alpha}{8}=5 \& \frac{\alpha}{4}=2$
$ \alpha=8$
image
$\text { Area }=\frac{1}{2}(6)(8)=24$
Perimeter $=24$
Circumradius $=5$
Inradius $=\frac{\Delta}{ s }=\frac{24}{12}=2$