The area of a triangle whose vertices are (−3, 0), (3, 0) and (0, k) is 6. Then the value (or the values) of k will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2, −2
By the determinant formula for the area of a triangle:
\(\displaystyle \Delta = \frac{1}{2}\begin{vmatrix}-3 & 0 & 1\\ 3 & 0 & 1\\ 0 & k & 1\end{vmatrix}\)
Expanding along the second column (only the entry k is non-zero):
\(\displaystyle \Delta = \frac{1}{2}\left[-k\,(-3-3)\right] = 3k\)
The area is the absolute value, so
\(|3k| = 6\)
\(k = 2 \text{ or } k = -2\)
Check: the base from (−3, 0) to (3, 0) is 6 and the height is |k| = 2, so the area is \(\frac{1}{2} \times 6 \times 2 = 6\) for both values.
Why the other options are incorrect
Option (2) 2 — Incorrect. k = −2 also gives area 6, so this misses a value.
Option (3) −2 — Incorrect. k = 2 also gives area 6, so this misses a value.
Option (4) 1 — Incorrect. With k = 1 the area is \(\frac{1}{2} \times 6 \times 1 = 3\).