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प्रत्येक में $k$ का मान ज्ञात कीजिए यदि त्रिभुजों का क्षेत्रफल $4$ वर्ग इकाई है जहाँ शीर्षबिंदु निम्नलिखित हैं:
$(\mathrm{k}, 0),(4,0),(0,2)$
$0,3$
$0,5$
$0,8$
$0,9$
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 $(k, 0),(4,0),(0,2)$ is given by the relation,
$\Delta=\frac{1}{2}\left|\begin{array}{lll}k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1\end{array}\right|$
$=\frac{1}{2}[k(0-2)-0(4-0)+1(8-0)]$
$=\frac{1}{2}[-2 k+8]=k+4$
$\therefore-k+4=\pm 4$
When $-k+4=-4, k=8$
When $-k+4=-4, k=0$
Hence, $k=0,8$
