If the area of the triangle formed by the points (x, 2x),

We have the co-ordinates of the vertices of the triangle as $\mathrm{A}(x, 2 x) ; \mathrm{B}(-2,6) ; \mathrm{C}(3,1)$ which has an area of 5 sq.units.

In general if $\mathrm{A}\left(x_{1}, y_{1}\right) ; \mathrm{B}\left(x_{2}, y_{2}\right) ; \mathrm{C}\left(x_{3}, y_{3}\right)$ are non-collinear points then area of the triangle formed is given by-,

$\operatorname{ar}(\Delta \mathrm{ABC})=\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

So,

$5=\frac{1}{2}|x(6-1)-2(1-2 x)+3(2 x-6)|$

$5=\frac{1}{2}|15 x-20|$

Simplify the modulus function to get,

$3 x-4=\pm 2$

$x=\frac{4 \pm 2}{3}$

Therefore,

$x=2, \frac{2}{3}$

So the answer is (a)