Find area of the triangle with vertices at the point given in each of the following: $(2,7),(1,1),(10,8)$
$\frac{43}{2}$ square units
$\frac{45}{2}$ square units
$\frac{49}{2}$ square units
$\frac{47}{2}$ square units
Find the equation of the line joining $\mathrm{A}(1,3)$ and $\mathrm{B}(0,0)$ using determinants and find $\mathrm{k}$ if $\mathrm{D}(\mathrm{k}, 0)$ is a point such that area of triangle $\mathrm{ABD}$ is $3 \,\mathrm{sq}$ $\mathrm{units}$.
Let $S$ be the set of all $\lambda \in \mathrm{R}$ for which the system of linear equations
$2 x-y+2 z=2$
$x-2 y+\lambda z=-4$
$x+\lambda y+z=4$
has no solution. Then the set $S$
Let the system of linear equations
$x+y+\alpha z=2$
$3 x+y+z=4$
$x+2 z=1$
have a unique solution $\left(x^{*}, y^{*}, z^{*}\right)$. If $\left(\alpha, x^{*}\right),\left(y^{*}, \alpha\right)$ and $\left(x^{*},-y^{*}\right)$ are collinear points, then the sum of absolute values of all possible values of $\alpha$ is
The value of $k \in R$, for which the following system of linear equations
$3 x-y+4 z=3$
$x+2 y-3 x=-2$
$6 x+5 y+k z=-3$
has infinitely many solutions, is:
Let $\omega = - \frac{1}{2} + i\frac{{\sqrt 3 }}{2}$. Then the value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{ - 1 - {\omega ^2}}&{{\omega ^2}}\\1&{{\omega ^2}}&{{\omega ^4}}\end{array}\,} \right|$ is