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, \(k = 0, 8.\)