Question

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

Answer 1

We have,

$\therefore$ Area of $△DBC=\frac{1}{2}\left|\right(5x+6+12x)-(-9x+20-2x\left)\right|$

$\Rightarrow$ Area of $△DBC=\frac{1}{2}\left|\right(28x-14\left)\right|=|14x-7|=7|2x-1|$

Also, we have

$\therefore$ Area of $△ABC$ $=\frac{1}{2}\left|\right(30+6+12)-(-9+20-12\left)\right|$

$\Rightarrow$ Area of $△ABC$= $\frac{1}{2}|48+1|=\frac{49}{2}$

Now, $\frac{Areaof△DBC}{Areaof△ABC}=\frac{1}{2}$

$\Rightarrow$ $\frac{7|2x-1|}{\frac{49}{2}}=\frac{1}{2}$

$\Rightarrow$ $|2x-1|=\frac{7}{4}$

$\Rightarrow$ 2x- 1 = $\ne \frac{7}{4}$

$\Rightarrow$ $2x=\frac{11}{4}or2x=-\frac{3}{4}$

$\Rightarrow$ $x=\frac{11}{8}orx=-\frac{3}{8}$