Question

The area of a triangle with vertices \((3, 0), (0, k), (-3, 0)\) is \(9\) sq units. The value of \(k\) is:

• A. \(9\)
• B. \(-9\)
• C. \(3\)
• D. \(6\)

Hint:

To find the area of a triangle using coordinate geometry, always apply the determinant or formula-based method and take the absolute value to ensure the area remains positive.

Answer 1

The Correct Option is C

We are given three vertices of a triangle: \[ A(3, 0),\quad B(0, k),\quad C(-3, 0) \] The area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is given by: \[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \] Substituting the given values: \[ = \frac{1}{2} \left| 3(k - 0) + 0(0 - 0) + (-3)(0 - k) \right| = \frac{1}{2} \left| 3k + 0 + 3k \right| = \frac{1}{2} \left| 6k \right| \] Given that the area is \(9\) square units: \[ \frac{1}{2} |6k| = 9 \Rightarrow |6k| = 18 \Rightarrow k = \pm3 \] Among the given options, only \(3\) is present. \[ \boxed{k = 3} \]