Question

Four points $A(6,3),B(-3,5),C(4,-2)$ and $D(x,3x)$ are given such that $\frac{△DBC}{△ABC}=\frac{1}{2}$, find $x$.

Answer 1

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

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

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

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

$=14x-7$

Area of $△ABC=\frac{1}{2}\left[6\right(5+2)-3(-2-3)+4(3-5\left)\right]$

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

$=\frac{49}{2}$

According to question,

$\frac{△DBC}{△ABC}=\frac{1}{2}$

$\frac{14x-7}{49/2}=\frac{1}{2}$

$14x-7=\frac{49}{4}$

$14x=\frac{77}{4}$

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