Question

If the area of \( \triangle PQR \) with vertices \( P(k, 1), Q(2, 4), R(1, 1) \) is 3 square units, find \( k \).

Hint:

To find the area of a triangle given its vertices, use the determinant-like formula \( \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \).

Answer 1

Step 1: The area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is given by the formula: \[ {Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|. \] Substituting the coordinates \( P(k, 1), Q(2, 4), R(1, 1) \) into this formula: \[ {Area} = \frac{1}{2} \left| k(4 - 1) + 2(1 - 1) + 1(1 - 4) \right|. \] Step 2: Simplifying the expression: \[ {Area} = \frac{1}{2} \left| k(3) + 0 + 1(-3) \right| = \frac{1}{2} \left| 3k - 3 \right|. \] Step 3: We are given that the area is 3 square units, so: \[ \frac{1}{2} \left| 3k - 3 \right| = 3 \quad \Rightarrow \quad \left| 3k - 3 \right| = 6. \] Step 4: Solving the absolute value equation: \[ 3k - 3 = 6 \quad {or} \quad 3k - 3 = -6. \] For \( 3k - 3 = 6 \): \[ 3k = 9 \quad \Rightarrow \quad k = 3. \] For \( 3k - 3 = -6 \): \[ 3k = -3 \quad \Rightarrow \quad k = -1. \] Step 5: Therefore, the possible values of \( k \) are \( k = -1 \) and \( k = 3 \).