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If $\left (-4, 0 \right), \left(1, -1 \right)$ are two vertices of a triangle whose area is $4$ Sq units then its third vertex lies on :

  1. $y=x$
  2. $5x+y+12=0$
  3. $x+5y-4=0$
  4. $x-5y+4=0$
in Quantitative Aptitude 9.3k points 41 648 824
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1 Answer

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$\begin{align} \Delta &=\frac{1}{2} \left| \begin{matrix} x_1 & y_1 & 1\cr x_2 & y_2 & 1 \cr x_3 & y_3 & 1 \cr \end{matrix} \right| \\  &= \frac{1}{2} [ x_1 (y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2)] = 4 \\ & \implies  x_1 (y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2) = 8 \\ \\ &\text{put } (x_1,y_1), (x_2,y_2),(x_3,y_3) \rightarrow (x,y), (-4,0), (1,-1) \text{ respectively,}\\ \\ &\implies  x (0-(-1)) + (-4) ((-1) – y) + 1 (y-0) = 8 \\ &\implies x + 4y +4+y =8 \\ & \implies x+5y -4 =0 \end{align} $

Option C.

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