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If $(-4,0),(1,-1)$ 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. $x-5y+4=0$
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Answer is C
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Answer is C.

Area = $\frac{1}{2} determinant\begin{vmatrix} -4&0&1 \\ 1&-1&1 \\ x&y&1 \end{vmatrix} = 4$

After expanding, we will get x + 5y – 4 =0

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