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જો $(-2,0),(0,4),(0, \mathrm{k})$ શિરોબિંદુવાળા ત્રિકોણનું ક્ષેત્રફળ $4$ ચોરસ એકમ હોય, તો $\mathrm{k}$ નું મૂલ્ય શોધો.
$0,5$
$0,7$
$0,2$
$0,8$
Solution
We Know that the area of a triangle whose vertices are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ is the absolute value of the determinant ( $\Delta$ ), where
$\Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{array}\right|$
It is given that the area of triangle is $4$ square units.
$\therefore \Delta=\pm 4$
The area of the triangle with vertices $(-2,0),(0,4),(0, k)$ is given by the relation,
$\Delta=\frac{1}{2}\left|\begin{array}{ccc}-2 & 0 & 1 \\ 0 & 4 & 1 \\ 0 & k & 1\end{array}\right|$
$=\frac{1}{2}[-2(4-k)]$
$=k-4$
$\therefore k-4=\pm 4$
When $k-4=-4, k=0$
When $k-4=4, k=8$
Hence, $k=0,8$
