Question

The coordinate of points $A,B,C$ and $D$ are $(-3,5),(4,-2),(x,3x)$ and $(6,3)$ respectively and $\frac{△ABC}{△BCD}=\frac{2}{3}$, find $x$

Answer 1

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

Area of $△$ $ABC=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$

$=\frac{1}{2}[-3(-2-3x)+4(3x-5)+x(5+2\left)\right]$

$=\frac{1}{2}[6+9x+12x-20+5x+2x]=\frac{28x-14}{2}=14x-7$

Area of $△$ $BCD=\frac{1}{2}\left[4\right(3x-3)+x(3+2)+6(-2-3x\left)\right]$

$=\frac{1}{2}[12x-12+3x+2x-12-18x]$

$=\frac{-x-24}{2}$

$\frac{△ABC}{△BCD}=\frac{2}{3}\Rightarrow \frac{14x-7}{\frac{-x-24}{2}}=\frac{2}{3}\Rightarrow 3(14x-7)=2\times \frac{-x-24}{2}$

$42x-21=-x-24\Rightarrow 42x-21+x+24=0$

$43x+3=0$

$x=-\frac{3}{43}$