If the area of the triangle with vertices \((k, 0)\), \((6, 0)\) and \((0, 3)\) is 6 square units, then
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(k = 2\) or \(k = 10\)
The area is the absolute value of the determinant expression:
\( \displaystyle \frac{1}{2}\begin{vmatrix}k & 0 & 1\\ 6 & 0 & 1\\ 0 & 3 & 1\end{vmatrix} = \pm 6 \)
Expanding along the first row:
\(k(0 - 3) - 0(6 - 0) + 1(18 - 0) = \pm 12\)
\(18 - 3k = \pm 12\)
Taking the two signs:
\(18 - 3k = 12\) gives \(k = 2\)
\(18 - 3k = -12\) gives \(k = 10\)
Both values work: the base on the x-axis has length \(|6 - k| = 4\) and the height is 3, so the area is \(\frac{1}{2} \times 4 \times 3 = 6\).
Why the other options are incorrect
Option (2) \(k = 10\) only — Incorrect. This takes only the minus sign; \(k = 2\) also gives area \(\frac{1}{2}|18 - 6| = 6\).
Option (3) \(k = -2\) or \(k = 10\) — Incorrect. With \(k = -2\) the area is \(\frac{1}{2}|18 + 6| = 12\).
Option (4) \(k = 2\) or \(k = -10\) — Incorrect. With \(k = -10\) the area is \(\frac{1}{2}|18 + 30| = 24\).