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3 and 4 .Determinants and Matrices
medium
ધારો કે $A (1, \alpha)$, $B (\alpha, 0)$ અને $C (0, \alpha)$ શિરોબિંદુઆવાળા ત્રિકોણનું ક્ષેત્રફળ $4$ ચોરસ એકમ છે. જો બિંદુઆ $(\alpha,-\alpha),(-\alpha, \alpha)$ અને $\left(\alpha^{2}, \beta\right)$ સમરેખ હોય, તો $\beta$ =...........
A
$64$
B
$-8$
C
$-64$
D
$512$
(JEE MAIN-2022)
Solution
$\frac{1}{2}\left|\begin{array}{lll}\alpha & 0 & 1 \\ 1 & \alpha & 1 \\ 0 & \alpha & 1\end{array}\right|=\pm 4$
$\alpha=\pm 8$
Now given points $(8,-8),(-8,8),(64, \beta)$
$OR (-8,8),(8,-8),(64, \beta)$
are collinear $\Rightarrow$ Slope $=-1$.
$\beta=-64$
Standard 12
Mathematics
