Question

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

• A. 6
• B. 9
• C. 3
• D. -9

Hint:

To find the area of a triangle given its vertices, use the determinant formula for the area. The formula \( \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \) is useful when you have the coordinates of the vertices. Always ensure to use the correct coordinates in the formula.

Answer 1

The Correct Option is C

The correct answer is (C) : 3.

Answer 2

The correct answer is: (C) 3.

We are given the coordinates of the vertices of the triangle: (-3, 0), (3, 0), and (0, k). The area of the triangle is given as 9 square units, and we are tasked with finding the value of \( k \).

The formula for the area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is:

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

Substituting the coordinates of the vertices (-3, 0), (3, 0), and (0, k) into the formula:

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

Simplifying the expression:

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

\( \text{Area} = \frac{1}{2} \left| 3k + 3k \right| = \frac{1}{2} \left| 6k \right| \)

We are given that the area is 9 square units, so:

\( \frac{1}{2} \left| 6k \right| = 9 \)

Multiplying both sides by 2:

\( \left| 6k \right| = 18 \)

Therefore, \( 6k = 18 \) or \( 6k = -18 \). Solving for \( k \):

\( k = 3 \) or \( k = -3 \)

Since the vertex (0, k) is above the x-axis, we take \( k = 3 \). Thus, the value of \( k \) is (C) 3.