If $\mathrm{A}(6, 3), \mathrm{B}(-3, 5), \mathrm{C}(4,-2), \mathrm{D}(\mathrm{x}, 3\mathrm{x})$ are four points and the magnitude of the area of $∆\mathrm{ABC}$ is twice the area of $∆\mathrm{DCB}$ then x=

Given points A=(6, 3), B=(-3, 5), C=(4, -2), D=(x, 3x)

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

$$\begin{gathered} = \frac{1}{2}\left|42+15-8\right| \\ =\frac{1}{2}\left|49\right| \\ =\frac{49}{2} \end{gathered}$$

 Area of $∆ \mathrm{DCB}=\frac{1}{2}\left|\mathrm{x}(-2-5)+4(5-3x)+(-3)(3\mathrm{x}+2)\right|$

 $$\begin{gathered} =\frac{1}{2}\left|-7\mathrm{x}+20-12\mathrm{x}-9\mathrm{x}-6\right| \\ =\frac{1}{2}\left|-28\mathrm{x}+14\right| \\ = \left|-14\mathrm{x}+7\right| \end{gathered}$$

Given that Area of $∆ \mathrm{ABC}=2 (\mathrm{Area} \mathrm{of} ∆ \mathrm{DCB})$

$$\begin{gathered} \Rightarrow \frac{49}{2}=2\left|-14\mathrm{x}+7\right| \\ \Rightarrow 49=4\times \left|7(-2\mathrm{x}+1)\right| \\ \Rightarrow 7=4\left|-2\mathrm{x}+1\right| \\ \Rightarrow -2\mathrm{x}+1=\pm \frac{7}{4} \\ \Rightarrow -2\mathrm{x}+1=\frac{7}{4} \left(\mathrm{or}\right) -2\mathrm{x}+1=\frac{-7}{4} \\ \Rightarrow -2\mathrm{x}=\frac{7}{4}-1 \left(\mathrm{or}\right) -2\mathrm{x}=\frac{-7}{4}-1 \\ \Rightarrow -2\mathrm{x}=\frac{3}{4} -2\mathrm{x}=\frac{-11}{4} \\ \Rightarrow \mathrm{x}=\frac{-3}{8} \Rightarrow \mathrm{x}=\frac{11}{8} \\ \therefore \mathrm{x}=\frac{-3}{8} \left(\mathrm{or}\right) \frac{11}{8} \end{gathered}$$