If the area of a triangle with vertices (-3, 0), (3, 0) and (0, k) is 9 sq. units, then k equals

±3

The correct answer is Option (3) → ±3

Let the vertices of the triangle be:

$A(-3, 0),\ B(3, 0),\ C(0, k)$

Using the formula for area of triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$:

$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

Substitute values:

$= \frac{1}{2} \left| (-3)(0 - k) + (3)(k - 0) + (0)(0 - 0) \right|$

$= \frac{1}{2} \left| 3k + 3k \right| = \frac{1}{2} \cdot 6|k| = 3|k|$

Given area = 9

$3|k| = 9 \Rightarrow |k| = 3$

Therefore, $k = \pm 3$