If the area of a triangle with vertices (−3, 0), (3, 0) and (0, k) is 9 sq. units, then k equals
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\pm 3\)
Note: The given answer is as per the official NTA Answer Key. However, more than one answer is possible: Option (2), Option (3) and Option (4) are all correct, as explained below.
Area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\):
\( \displaystyle \Delta = \frac{1}{2}\left|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right| \)
\( \displaystyle = \frac{1}{2}\left|(-3)(0 - k) + 3(k - 0) + 0\right| = \frac{1}{2}|6k| = 3|k| \)
\(3|k| = 9\)
\(|k| = 3,\ \ k = \pm 3\)
Option (1) 0 — Incorrect. With k = 0 all three vertices lie on the x-axis and the area is 0.
Option (2) −3 — Should also be correct, though not the answer given by NTA. Area \(= 3|-3| = 9\).
Option (3) \(\pm 3\) — Correct. It gives both values of k for which the area is 9.
Option (4) 3 — Should also be correct, though not the answer given by NTA. Area \(= 3|3| = 9\).
Shortcut: The base from (−3, 0) to (3, 0) has length 6 and the height is \(|k|\). So \(\frac{1}{2} \times 6 \times |k| = 9\) gives \(|k| = 3\).
Advice for the exam: When more than one option is correct, mark the one that gives all the values, here ±3. If NTA’s provisional answer key marks a different option, challenge it during the answer-key challenge window, giving proper reasoning. NTA does revise its key when a challenge is well justified, so a sound challenge can raise your marks.