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  4. A(6, 3), B(- 3, 5), C(4, - 2) and D(x, 3x) are four points. If ΔDBC: ΔABC = 1: 2, then x is equal to

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Q. A(6, 3), B(- 3, 5), C(4, - 2) and D(x, 3x) are four points. If $\Delta$DBC: $\Delta$ABC = 1: 2, then x is equal to

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Solution:

$\Delta DBC = \frac{1}{2}| x\left(5 +2\right) 3\left( -2 +3x\right)+ 4\left(3x- 5\right) | $
$= \frac{1}{2} \left|28 x -14 \right| = \left|14x - 7\right|$
$ \Delta ABC = \frac{1}{2} \left|6\left(5+ 2\right) 3\left( 2+ 3\right) 4\left(3+ 5\right) \right| $
$= \frac{49}{2} $
Now $\left|14x -7\right| = \frac{1}{2} \times\frac{49}{2} \Rightarrow x = \frac{11}{8} $ or $- \frac{3}{8} $