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:

To find the area of a triangle given its vertices, use the determinant formula. The area is half the absolute value of the determinant of the matrix formed by the coordinates.

Answer 1

Step 1: Formula for the area of a triangle using determinants.
The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by the determinant formula: \[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1
x_2 & y_2 & 1
x_3 & y_3 & 1 \end{vmatrix} \right| \] Substitute the coordinates of the points \( (-3, 6) \), \( (-4, 4) \), and \( (k, -2) \): \[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} -3 & 6 & 1
-4 & 4 & 1
k & -2 & 1 \end{vmatrix} \right| \] Step 2: Calculating the determinant.
Now, calculate the determinant: \[ \begin{vmatrix} -3 & 6 & 1
-4 & 4 & 1
k & -2 & 1 \end{vmatrix} = -3 \begin{vmatrix} 4 & 1
-2 & 1 \end{vmatrix} - 6 \begin{vmatrix} -4 & 1
k & 1 \end{vmatrix} + 1 \begin{vmatrix} -4 & 4
k & -2 \end{vmatrix} \] Simplifying each 2x2 determinant: \[ \begin{vmatrix} 4 & 1
-2 & 1 \end{vmatrix} = 4(1) - 1(-2) = 4 + 2 = 6 \] \[ \begin{vmatrix} -4 & 1
k & 1 \end{vmatrix} = -4(1) - 1(k) = -4 - k \] \[ \begin{vmatrix} -4 & 4
k & -2 \end{vmatrix} = -4(-2) - 4(k) = 8 - 4k \] Substitute these into the determinant formula: \[ \text{Area} = \frac{1}{2} \left| -3(6) - 6(-4 - k) + 1(8 - 4k) \right| \] Simplify: \[ \text{Area} = \frac{1}{2} \left| -18 + 24 + 6k + 8 - 4k \right| \] \[ = \frac{1}{2} \left| 14 + 2k \right| \] We are given that the area is 12 square units, so: \[ \frac{1}{2} \left| 14 + 2k \right| = 12 \] Multiply both sides by 2: \[ \left| 14 + 2k \right| = 24 \] This gives two cases: \[ 14 + 2k = 24 \quad \text{or} \quad 14 + 2k = -24 \] Step 3: Solving for \( k \).
Solving \( 14 + 2k = 24 \): \[ 2k = 10 \quad \Rightarrow \quad k = 5 \] Solving \( 14 + 2k = -24 \): \[ 2k = -38 \quad \Rightarrow \quad k = -19 \] Thus, the possible values of \( k \) are 5 and -19. Step 4: Conclusion.
The value of \( k \) is either 5 or -19.