if A6,3,B−3,5,C4,−2 and Dx,3x are four points if arΔDBCarΔABC=1:2then x is equal to

A(6,3),B(−3,5),C(4,−2),D(x,3x)[DBC]=12x3x1−3514−21=12|7x−3x(−7)+1(−14)|=12|28x−14|=|14x−7|[ABC]=12631−3514−21=12|6(7)−3(−7)+1(−14)|=492[DBC][ABC]=12|14x−7|492=12|2x−1|=742x−1=74,−742x=114,−34x=118,−38

if A6,3,B−3,5,C4,−2 and Dx,3x are four points if arΔDBCarΔABC=1:2then x is equal to

A(6,3),B(−3,5),C(4,−2),D(x,3x)[DBC]=12x3x1−3514−21=12|7x−3x(−7)+1(−14)|=12|28x−14|=|14x−7|[ABC]=12631−3514−21=12|6(7)−3(−7)+1(−14)|=492[DBC][ABC]=12|14x−7|492=12|2x−1|=742x−1=74,−742x=114,−34x=118,−38

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if $A (6, 3), B (- 3, 5), C (4, - 2)$ and $D (x, 3 x)$ are four points if $\frac{ar (ΔDBC)}{ar (ΔABC)} = 1 : 2$ then x is equal to

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$\frac{11}{8}$

$\frac{8}{11}$

answer is A.

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Detailed Solution

$A (6, 3), B (- 3, 5), C (4, - 2), D (x, 3 x)$
$\begin{array}{l} [DBC] = \frac{1}{2} |\begin{array}{lcc} x & 3 x & 1 \\ - 3 & 5 & 1 \\ 4 & - 2 & 1 \end{array}| \\ = \frac{1}{2} | 7 x - 3 x (- 7) + 1 (- 14) | \\ = \frac{1}{2} | 28 x - 14 | \\ = | 14 x - 7 | \\ [ABC] = \frac{1}{2} |\begin{array}{ccc} 6 & 3 & 1 \\ - 3 & 5 & 1 \\ 4 & - 2 & 1 \end{array}| \end{array}$

$\begin{array}{l} = \frac{1}{2} | 6 (7) - 3 (- 7) + 1 (- 14) | = \frac{49}{2} \\ \frac{[D B C]}{[A B C]} = \frac{1}{2} \\ \frac{| 14 x - 7 |}{\frac{49}{2}} = \frac{1}{2} \\ | 2 x - 1 | = \frac{7}{4} \\ 2 x - 1 = \frac{7}{4}, \frac{- 7}{4} \\ 2 x = \frac{11}{4}, \frac{- 3}{4} \\ x = \frac{11}{8}, \frac{- 3}{8} \end{array}$