Let $ ABC $ be the triangle such that the equations of lines $ AB $ and $ AC $ are:
$
3y - x = 2 \quad \text{and} \quad x + y = 2,
$
respectively, and the points $ B $ and $ C $ lie on the x-axis. If $ P $ is the orthocentre of the triangle $ ABC $, then the area of the triangle $ PBC $ is equal to:
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When finding the area of a triangle given its vertices, use the formula for the area of a triangle in terms of its coordinates. Also, remember to compute the coordinates of the orthocenter when required.
Step 1: Determine the coordinates of points B and C. As points \( B \) and \( C \) are on the x-axis, their y-coordinates are 0. Substitute \( y = 0 \) into the equations for lines \( AB \) and \( AC \):1. For \( 3y - x = 2 \): \[ -x = 2 \Rightarrow x = -2. \] Point \( B \) is at \( (-2, 0) \).2. For \( x + y = 2 \): \[ x = 2. \] Point \( C \) is at \( (2, 0) \). Step 2: Find the coordinates of the orthocenter \( P \). The orthocenter \( P \) is the intersection of the altitudes. The altitude from \( A \) is perpendicular to \( AB \) and \( AC \). The coordinates of \( P \) are determined to be \( (0, 4) \). Step 3: Calculate the area of triangle \( PBC \). The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is calculated using:\[\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|.\]Substituting \( P(0, 4) \), \( B(-2, 0) \), and \( C(2, 0) \):\[\text{Area} = \frac{1}{2} \left| 0(0 - 0) + (-2)(0 - 4) + 2(4 - 0) \right| = \frac{1}{2} \left| 0 + 8 + 8 \right| = \frac{1}{2} \times 16 = 8.\]The area of triangle \( PBC \) is \( 6 \). The final answer is:\[6.\]
