The least positive value of t,so that the lin | vedclass

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(D) The given lines are:$x - (t + \alpha) = 0$$y + 16 = 0$$-\alpha x + y = 0$Since these lines are concurrent,the determinant of the coefficients must

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$8$

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(D) The given lines are:$x - (t + \alpha) = 0$$y + 16 = 0$$-\alpha x + y = 0$Since these lines are concurrent,the determinant of the coefficients must be zero:$\begin{vmatrix} 1 & 0 & -(t+\alpha) \ 0 & 1 & 16 \ -\alpha & 1 & 0 \end{vmatrix} = 0$Expanding the determinant along the first row:$1(0 - 16) - 0 + (-(t + \alpha))(0 - (-\alpha)) = 0$$-16 - (t + \alpha)(\alpha) = 0$$-16 - t\alpha - \alpha^2 = 0$$\alpha^2 + t\alpha + 16 = 0$For $\alpha$ to be a real number,the discriminant $D$ must be greater than or equal to zero:$D = t^2 - 4(1)(16) \geq 0$$t^2 - 64 \geq 0$$t^2 \geq 64$Since $t$ must be positive,$t \geq 8$.The least positive value of $t$ is $8$.

The least positive value of $t$,so that the lines $x=t+\alpha, y+16=0$ and $y=\alpha x$ are concurrent,is

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