The area of $\triangle PQR$ with the vertices $P(k, 1)$,$Q(2, 4)$,and $R(1, 1)$ is $3$ sq. units. Then,$k = $ . . . . . . .

  • A
    $0, 2$
  • B
    $-3, 1$
  • C
    $-1, 3$
  • D
    $1, 3$

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Similar Questions

If $1, \omega, \omega^2$ are the cube roots of unity,then $\Delta = \begin{vmatrix} 1 & \omega^n & \omega^{2n} \\ \omega^n & \omega^{2n} & 1 \\ \omega^{2n} & 1 & \omega^n \end{vmatrix} = $

Let $S = \left\{ A = \begin{bmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{bmatrix} : a, b, c, d, e \in \{0, 1\} \text{ and } |A| \in \{-1, 1\} \right\}$,where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is:

If $a, b, c$ are distinct and rational numbers,then the value of the determinant $\left| \begin{array}{ccc} (a^2 + b^2 + c^2) & (ab + bc + ca) & (ab + bc + ca) \\ (ab + bc + ca) & (a^2 + b^2 + c^2) & (ab + bc + ca) \\ (ab + bc + ca) & (ab + bc + ca) & (a^2 + b^2 + c^2) \end{array} \right|$ is always:

The value of $a$ for which the system of equations has a non-zero solution is
$a^{3} x+(a+1)^{3} y+(a+2)^{3} z=0$
$a x+(a+1) y+(a+2) z=0$
$x+y+z=0$

$\Delta = \left| \begin{array}{ccc} a & a+b & a+b+c \\ 3a & 4a+3b & 5a+4b+3c \\ 6a & 9a+6b & 11a+9b+6c \end{array} \right|$ where $a = i, b = \omega, c = \omega^2$,then $\Delta$ is equal to

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