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# The area of a triangle with vertices(-3,0),(3,0) and (0,k) is 9 sq. units.The value of k will be$\begin{array}{1 1}(A)\quad 9 & (B)\quad 3\\(C)\quad-9 & (D)\quad 6\end{array}$

Toolbox:
• Area of a triangle is given by
• $\Delta=\frac{1}{2}\begin{vmatrix}x_1 & y_1 &1\\x_2 & y_2 &1\\x_3 & y_3 &1\end{vmatrix}$
Given the vertices of the $\Delta ABC$ are A(-3,0),B(3,0),C(0,k).
The area of the $\Delta ABC$=9sq.units
The area $\Delta=\begin{vmatrix}x_1 & y_1 & 1\\x_2 & y_2 &1\\x_3 & y_3 & 1\end{vmatrix}$
substituting for $(x_1,y_1),(x_2,y_2),(x_3,y_3)$
Where A(-3,0),B(3,0),C(0,k)
Therefore $\mid\Delta\mid=\frac{1}{2}\begin{vmatrix}-3 & 0 & 1\\3 & 0 & 1\\0 &k & 1\end{vmatrix}=9$
Expanding along $R_1$,
$\Delta=\frac{1}{2}\mid-3(0-k)-0+1(3k)\mid$
$9=\frac{1}{2}\mid3k+3k|$.
$\;\;=\frac{6k}{2}=9$
k=3
Hence B is the correct answer.