The coordinates of the points A, B, C, D are $(6,3),(-3,5),\text{}(4,-2)\text{and}(x,3x)$ respectively. If the area of the $\Delta ABC$is twice that of the area of $\Delta DBC$, then the value of $x$can be

$A(6,3),B(-3,5),\text{}C(4,-2)$

Area of $\Delta ABC=\frac{1}{2}|6(5+2)+(-3)(-2-3)+4(3-5)|$

$=\frac{1}{2}|6\left(7\right)-3(-5)+4(-2)|$

$=\frac{1}{2}|42+15-8|$

$=\frac{1}{2}\left|49\right|$

Area of $\Delta DBC$,

$D(x,3x),B(-3,5),C(4,-2)$

Area of $\Delta DBC=\frac{1}{2}|x(5+2)-3(-2-3x)+4(3x-5)|$

$=\frac{1}{2}|7x+6+9x+12x-20|$

$=\frac{1}{2}|28x-14|$

$=|14x-7|$

Area of $\Delta ABC=$ 2 Area of $\Delta DBC$

$\Rightarrow \frac{\left|49\right|}{2}=2|14x-7|$

$\Rightarrow 49=4(14x-7)$

$\Rightarrow 49=56x-28$

$\Rightarrow 49+28=56x$

$\Rightarrow 49+28=56x$

$\Rightarrow \frac{77}{56}=x$

$\therefore x=\frac{11}{8}$