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

Solution

✅ Correct Option: 3

The triangle has vertices at (−3,0)(-3, 0), (3,0)(3, 0), and (0,k)(0, k).

Points (−3,0)(-3, 0) and (3,0)(3, 0) both lie on the x-axis.

Point (0,k)(0, k) lies on the y-axis.


The base of the triangle is the distance from (−3,0)(-3, 0) to (3,0)(3, 0):

Base =3−(−3)= 3 - (-3)

Base =6= 6 units


The height is the perpendicular distance from (0,k)(0, k) to the x-axis.

Height =∣k∣= |k| units


Using the area formula:

Area=12×base×height\text{Area} = \dfrac{1}{2} \times \text{base} \times \text{height}

9=12×6×∣k∣9 = \dfrac{1}{2} \times 6 \times |k|

9=3∣k∣9 = 3|k|

∣k∣=3|k| = 3


Therefore, k=3k = 3 or k=−3k = -3

Both values give the same area since the formula uses absolute value.

The answer is k=±3k = \pm 3.

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