To find the ratio of the areas of triangles PBC and ABC given the points A(6, 3), B(3, 5), C(4, 2), and P(α, β), we will use the formula for the area of a triangle formed by three points in a 2D coordinate system. The formula for the area of a triangle with vertices at points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
### Step 1: Calculate the Area of Triangle ABC
Using the points A(6, 3), B(3, 5), and C(4, 2):
\[
\text{Area}_{ABC} = \frac{1}{2} \left| 6(5 - 2) + 3(2 - 3) + 4(3 - 5) \right|
\]
Calculating each term:
- \(6(5 - 2) = 6 \times 3 = 18\)
- \(3(2 - 3) = 3 \times (-1) = -3\)
- \(4(3 - 5) = 4 \times (-2) = -8\)
Now substituting these values into the area formula:
\[
\text{Area}_{ABC} = \frac{1}{2} \left| 18 - 3 - 8 \right| = \frac{1}{2} \left| 7 \right| = \frac{7}{2}
\]
### Step 2: Calculate the Area of Triangle PBC
Using the points P(α, β), B(3, 5), and C(4, 2):
\[
\text{Area}_{PBC} = \frac{1}{2} \left| \alpha(5 - 2) + 3(2 - \beta) + 4(\beta - 5) \right|
\]
Calculating each term:
- \(\alpha(5 - 2) = \alpha \times 3 = 3\alpha\)
- \(3(2 - \beta) = 3 \times (2 - \beta) = 6 - 3\beta\)
- \(4(\beta - 5) = 4\beta - 20\)
Now substituting these values into the area formula:
\[
\text{Area}_{PBC} = \frac{1}{2} \left| 3\alpha + (6 - 3\beta) + (4\beta - 20) \right|
\]
Simplifying:
\[
\text{Area}_{PBC} = \frac{1}{2} \left| 3\alpha + 6 - 3\beta + 4\beta - 20 \right| = \frac{1}{2} \left| 3\alpha + \beta - 14 \right|
\]
### Step 3: Calculate the Ratio of the Areas
Now we can find the ratio of the areas of triangles ABC and PBC:
\[
\text{Ratio} = \frac{\text{Area}_{ABC}}{\text{Area}_{PBC}} = \frac{\frac{7}{2}}{\frac{1}{2} \left| 3\alpha + \beta - 14 \right|} = \frac{7}{\left| 3\alpha + \beta - 14 \right|}
\]
### Final Answer
The ratio of the areas of triangles PBC and ABC is:
\[
\text{Ratio} = \frac{7}{3\alpha + \beta - 14}
\]
