Question

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

Answer 1

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

Area of $△DBC=\frac{1}{2}\left|\begin{array}{cc}-3& 5\\ 4& -2\\ x& 3x\\ -3& 5\end{array}\right|=\frac{1}{2}|-14+14x+14x|=7(2x-1)$

Area of $△ABC=\frac{1}{2}\left|\begin{array}{cc}6& 3\\ -3& 5\\ 4& -2\\ 6& 3\end{array}\right|=\frac{1}{2}\left(49\right)$

Given,$\frac{Areaof△DBC}{Areaof△ABC}=\frac{1}{2}=>\frac{7(2x-1)}{\frac{1}{2}\left(49\right)}=\frac{1}{2}=>56x-28=49=>x=\frac{11}{8}$