Key Concepts and Formulas

• Finding the intersection point of two lines by solving their equations simultaneously.
• The orthocenter of a triangle is the intersection point of its altitudes. The altitude from a vertex is perpendicular to the opposite side.
• If a line has slope $m$, a line perpendicular to it has slope $-1/m$.
• Area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Step-by-Step Solution

Step 1: Find the coordinates of points B and C.

Since B and C lie on the x-axis, their y-coordinates are 0. To find the x-coordinates, we substitute $y=0$ into the equations of lines AB and AC, respectively.

For point B (on line $3y - x = 2$): $3(0) - x = 2 \Rightarrow x = -2$. So, $B = (-2, 0)$.

For point C (on line $x + y = 2$): $x + 0 = 2 \Rightarrow x = 2$. So, $C = (2, 0)$.

Step 2: Find the coordinates of point A.

Point A is the intersection of lines AB and AC. We solve the system of equations: $3y - x = 2$ $x + y = 2$ From the second equation, $x = 2 - y$. Substituting this into the first equation: $3y - (2 - y) = 2 \Rightarrow 4y - 2 = 2 \Rightarrow 4y = 4 \Rightarrow y = 1$. Then, $x = 2 - y = 2 - 1 = 1$. So, $A = (1, 1)$.

Step 3: Find the equation of the altitude from A to BC.

Since BC lies on the x-axis (y=0), the altitude from A to BC is a vertical line passing through A. The equation of this line is $x = 1$.

Step 4: Find the equation of the altitude from C to AB.

The slope of line AB is found from its equation $3y - x = 2 \Rightarrow y = \frac{1}{3}x + \frac{2}{3}$. So, the slope of AB is $m_{AB} = \frac{1}{3}$.

The altitude from C to AB is perpendicular to AB, so its slope is $m = -\frac{1}{m_{AB}} = -3$. This altitude passes through the point C(2, 0). Using the point-slope form of a line, the equation of the altitude is: $y - 0 = -3(x - 2) \Rightarrow y = -3x + 6$.

Step 5: Find the coordinates of the orthocenter P.

The orthocenter P is the intersection of the altitudes from A to BC and from C to AB. We solve the system of equations: $x = 1$ $y = -3x + 6$ Substituting $x = 1$ into the second equation: $y = -3(1) + 6 = 3$. So, the orthocenter is $P = (1, 3)$.

Step 6: Calculate the area of triangle PBC.

The vertices of triangle PBC are $P(1, 3)$, $B(-2, 0)$, and $C(2, 0)$. Using the determinant formula for the area of a triangle: $\text{Area} = \frac{1}{2} |1(0 - 0) + (-2)(0 - 3) + 2(3 - 0)| = \frac{1}{2} |0 + 6 + 6| = \frac{1}{2} |12| = 6$ Alternatively, we can use the base-height formula. The base BC lies along the x-axis and has length $2 - (-2) = 4$. The height of the triangle is the y-coordinate of P, which is 3. So, the area of triangle PBC is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6$.

Common Mistakes & Tips

• Carefully solve the system of equations to find the intersection points. A small error can propagate through the entire solution.
• Remember that the altitude is perpendicular to the side, so use the negative reciprocal of the slope.
• When calculating the area, double-check the coordinates and the determinant formula to avoid errors.

Summary

We found the coordinates of vertices A, B, and C by finding the intersection of the given lines. Then, we determined the equations of two altitudes and found their intersection point, which is the orthocenter P. Finally, we calculated the area of triangle PBC using the coordinates of P, B, and C. The area of triangle PBC is 6.

Final Answer

The final answer is \boxed{6}, which corresponds to option (D).