Question

Find values of k if area of triangle is 4 square units and vertices are$(i)(k,0),(4,0),(0,2)(ii)(−2,0),(0,4),(0,k)$

Answer 1

We know that the area of a triangle whose vertices are $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is the absolute value of the determinant $(∆)$, where
$∆=\frac{1}{2}\begin{bmatrix}x_1&y_1&1\\\ x_2&y_2&1\\\ x_3&y_3&1\end{bmatrix}$
It is given that the area of triangle is 4 square units.
$∴∆ = ± 4.$
(i) The area of the triangle with vertices $(k, 0), (4, 0), (0, 2)$ is given by the relation,
$∆=\frac{1}{2}\begin{bmatrix}k&0&1\\\ 4&0&1\\\ 0&2&1\end{bmatrix}$
$=\frac{1}{2}[k(0-2)-0(4-0)+(8-0)]$
$=\frac{1}{2}[-2k+8]=-k+4$
$∴−k + 4 = ± 4$
When $−k + 4 = − 4, k = 8.$
When $−k + 4 = 4, k = 0.$
Hence, $k = 0, 8.$
(ii) The area of the triangle with vertices $(−2, 0), (0, 4), (0, k)$ is given by the relation,
$∆=\frac{1}{2}\begin{bmatrix}-2&0&1\\\ 0&4&1\\\ 0&k&1\end{bmatrix}$
$=\frac{1}{2}[-2(4-k)]$
$=k-4$
$∴k−4=±4$
When $k − 4 = − 4, k = 0$.
When $k − 4 = 4, k = 8.$
Hence, $k = 0, 8.$