Question

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

Answer 1

We know that the area of a triangle with vertices $\left(x_1,y_1\right),\left(x_2,y_2\right)\mathrm{and}\left(x_3,y_3\right)$ is given by:

$\text{Area}=\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\}$
$\therefore \mathrm{Area}\mathrm{of}∆DBC=\frac{1}{2}\left|-3\left(-2-3x\right)+4\left(3x-5\right)+x\left(5+2\right)\right|$
$\Rightarrow \mathrm{Area}\mathrm{of}∆DBC=7\left(2x-1\right)$

$\therefore \mathrm{Area}\mathrm{of}∆ABC=\frac{1}{2}\left|6\left(5+2\right)-3\left(-2-3\right)+4\left(3-5\right)\right|$
$=\frac{49}{2}$

It is given that $\frac{\mathrm{\Delta }DBC}{\mathrm{\Delta }ABC}=\frac{1}{2}$.
$\therefore \frac{7\left(2x-1\right)\times 2}{49}=\frac{1}{2}$

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