Question

Using determinants, find the value of \( k \) if the area of the triangle formed by the points \( (-3, 6), (-4, 4) \) and \( (k, -2) \) is 12 sq. units.

Hint:

When dealing with area problems involving determinants, always consider both the positive and negative results for the absolute value to find all possible values of the unknown coordinate.

Answer 1

Step 1: Understanding the Concept:
The area of a triangle with vertices \( (x_1, y_1), (x_2, y_2) \) and \( (x_3, y_3) \) is given by:
\[ \text{Area} = \frac{1}{2} | \Delta | \]
where \( \Delta \) is the determinant:
\[ \Delta = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]
Step 2: Detailed Explanation:
Given vertices: \( (-3, 6), (-4, 4), (k, -2) \) and Area \( = 12 \).
So, \( \frac{1}{2} | \Delta | = 12 \implies | \Delta | = 24 \).
\[ \Delta = \begin{vmatrix} -3 & 6 & 1 \\ -4 & 4 & 1 \\ k & -2 & 1 \end{vmatrix} \]
Expanding along the first row:
\[ \Delta = -3(4 - (-2)) - 6(-4 - k) + 1(8 - 4k) \]
\[ \Delta = -3(6) + 24 + 6k + 8 - 4k \]
\[ \Delta = -18 + 24 + 6k + 8 - 4k = 2k + 14 \]
Now, \( |2k + 14| = 24 \).
Case 1: \( 2k + 14 = 24 \implies 2k = 10 \implies k = 5 \).
Case 2: \( 2k + 14 = -24 \implies 2k = -38 \implies k = -19 \).
Step 3: Final Answer:
The values of \( k \) are 5 and \( -19 \).