Question

The coordinates of A,B,C and D are (6,3), (-3,5), (4,-2) and (k,3k) respectively. If $\frac{areaof\Delta DBC}{areaof\Delta ABC}=\frac{1}{2}$ , find k.

• A. $-\frac{5}{4}$
• B. $\frac{7}{4}$
• C. $\frac{5}{4}$
• D. $\frac{1}{4}$

Answer 1

Answer: (C)

$\frac{5}{4}$

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

Area of $△DBC=\left|\begin{array}{cc}-3& 5\\ 4& -2\\ k& 3k\\ -3& 5\end{array}\right|=\frac{1}{2}|6-20+12k+2k+5k+9k|=\frac{|14k-7|}{2}$

Area of $△ABC=\left|\begin{array}{cc}6& 3\\ -3& 5\\ 4& -2\\ 6& 3\end{array}\right|=\frac{1}{2}|30+9+6-20+12+12|=\frac{49}{2}$

Given $\frac{Areaof△DBC}{Areaof△ABC}=\frac{1}{2}=>\frac{14k-7}{49}=\frac{1}{2}=>2\left(14\right)k=49-14=>k=\frac{5}{4}$