If the lines \(\mathrm{x}=\mathrm{a}+\mathrm{m}, \mathrm{y}=-2\) and \(\mathrm{y}=\mathrm{mx}\) are concurrent, the least value of \(|\mathrm{a}|\) is \(\ldots \ldots .\) (a) 0 (b) \(\sqrt{2}\) (c) \(2 \sqrt{2}\) (d) none

To find the point of intersection for the lines \(\mathrm{x}=\mathrm{a}+\mathrm{m}\) and \(\mathrm{y}=-2\), we can substitute the value of x from the first equation, and use the second equation given. We have the following system of equations: \(\begin{cases} \mathrm{x}=\mathrm{a}+\mathrm{m} \\ \mathrm{y}=-2 \\ \end{cases}\) Since we're looking for the point of intersection, we'll need the coordinates (x, y). But first, let's find the value of x in terms of a and m: \(\mathrm{x}=\mathrm{a}+\mathrm{m}\) Therefore, the point of intersection between the first two lines is \((\mathrm{a}+\mathrm{m},-2)\).

To find the point of intersection for the lines \(\mathrm{y}=-2\) and \(\mathrm{y}=\mathrm{mx}\), we will substitute the value of y from the second equation into the third equation: \(\mathrm{y}=\mathrm{mx} \Rightarrow -2 = \mathrm{mx}\) Now, we need to express x in terms of m: \(x = -\frac{2}{\mathrm{m}}\) The point of intersection for the second and third lines is \(\left(-\frac{2}{\mathrm{m}}, -2\right)\).

Since the given lines are concurrent, the points of intersection we found in steps 1 and 2 must be the same point. So, we set the x-coordinates equal to each other: \(\mathrm{a}+\mathrm{m} = -\frac{2}{\mathrm{m}}\) Now, we solve for a: \(\mathrm{a} = -\frac{2}{\mathrm{m}} - \mathrm{m}\)

In order to minimize the value of \(|\mathrm{a}|\), we find the derivative of the function with respect to m, and set it equal to zero: \(\frac{\mathrm{d}|\mathrm{a}|}{\mathrm{d}\mathrm{m}} = \frac{\mathrm{d}|-2/\mathrm{m} - \mathrm{m}|}{\mathrm{d}\mathrm{m}} = 0\) Using calculus, we can determine that the minimum value for \(|\mathrm{a}|\) occurs at \(\mathrm{m} = \pm \sqrt{2}\). Plugging this value back into our expression for \(\mathrm{a}\), we get: \(a = -\frac{2}{\pm \sqrt{2}} - (\pm \sqrt{2})\) Plugging the positive value of \(\sqrt{2}\) into the equation, we obtain the least value of \(|\mathrm{a}|\): \(a = -\frac{2}{\sqrt{2}} - \sqrt{2}\) Thus, the least value of \(|\mathrm{a}|\) is \(\boxed{\sqrt{2}}\), which corresponds to option (b).