Question

If the area of \( \Delta ABC \) with vertices \( A(3, 1) \), \( B(-2, 1) \) and \( C(0, k) \) is \( 5 \) sq. units, then values of \( k \) are :

• A. \( 3, 1 \)
• B. \( -1, 3 \)
• C. \( -1, 2 \)
• D. \( 0, 2 \)

Hint:

Don't forget to include the \( \pm \) sign when removing the area modulus context. Omitting the negative case will cause you to miss one of the valid solution coordinates.

Answer 1

The Correct Option is B

Concept: The area of a triangle with coordinate vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be evaluated using determinants: \[ \text{Area} = \frac{1}{2} \left| \det \begin{bmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{bmatrix} \right| \] Because area is always a positive scalar, we must use an absolute value sign, which produces two possibilities (\( \pm \)) when solving for an unknown coordinate.

Step 1: Set up the determinant formula with given points. Substitute the vertices \( A(3, 1) \), \( B(-2, 1) \), and \( C(0, k) \) into the triangle formula: \[ \pm 5 = \frac{1}{2} \det \begin{bmatrix} 3 & 1 & 1 \\ -2 & 1 & 1 \\ 0 & k & 1 \end{bmatrix} \] Multiply both sides by \( 2 \) to clear the fraction: \[ \det \begin{bmatrix} 3 & 1 & 1 -2 & 1 & 1 0 & k & 1 \end{bmatrix} = \pm 10 \]

Step 2: Expand the matrix determinant. Let us expand along the first column to simplify calculations: \[ 3 \cdot \det \begin{bmatrix} 1 & 1 \\ k & 1 \end{bmatrix} - (-2) \cdot \det \begin{bmatrix} 1 & 1 \\ k & 1 \end{bmatrix} + 0 = \pm 10 \] \[ 3(1 - k) + 2(1 - k) = \pm 10 \] Combine the like terms together: \[ 5(1 - k) = \pm 10 \implies 1 - k = \frac{\pm 10}{5} \implies 1 - k = \pm 2 \]

Step 3: Solve the two separate equations for \( k \).
• Case 1 (Positive sign): \[ 1 - k = 2 \implies -k = 2 - 1 \implies -k = 1 \implies k = -1 \]
• Case 2 (Negative sign): \[ 1 - k = -2 \implies -k = -2 - 1 \implies -k = -3 \implies k = 3 \] Thus, the possible values for the coordinate parameter are \( k = -1 \) and \( k = 3 \).