If the area of the triangle with vertices \((1, 2)\), \((k, -1)\) and \((4, k)\) is \(\displaystyle\frac{15}{2}\) square units, then the sum of all possible real values of k is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
The area is the absolute value of
\( \displaystyle \frac{1}{2}\begin{vmatrix}1 & 2 & 1\\ k & -1 & 1\\ 4 & k & 1\end{vmatrix} \)
Expanding along the first row:
\(1(-1 - k) - 2(k - 4) + 1(k^{2} + 4)\)
\(= -1 - k - 2k + 8 + k^{2} + 4\)
\(= k^{2} - 3k + 11\)
This quadratic has discriminant \(9 - 44 \lt 0\), so it is always positive and the absolute value can be dropped:
\( \displaystyle \frac{1}{2}(k^{2} - 3k + 11) = \frac{15}{2} \)
\(k^{2} - 3k - 4 = 0\)
\((k - 4)(k + 1) = 0\)
So \(k = 4\) or \(k = -1\), and the sum is \(4 + (-1) = 3\).