Key Concepts and Formulas

• Section Formula: The coordinates of a point $P(x, y)$ dividing the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m:n$ are given by: $x = \frac{mx_2 + nx_1}{m+n}, \quad y = \frac{my_2 + ny_1}{m+n}$
• Area Ratio in Triangles: If two triangles share a common vertex and their bases lie on the same line, the ratio of their areas is equal to the ratio of their bases.
• Equation of a Line (Two-Point Form): The equation of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$
• Area of a Triangle: Area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$.

Step-by-Step Solution

Step 1: Finding the Ratio BP:PC

We are given that $\frac{\Delta_1}{\Delta_2} = \frac{4}{7}$, where $\Delta_1$ is the area of $\triangle APB$ and $\Delta_2$ is the area of $\triangle ABC$. Since triangles $APB$ and $ABC$ share the vertex $A$ and their bases $BP$ and $BC$ lie on the same line, the ratio of their areas is equal to the ratio of their bases. Therefore, $\frac{BP}{BC} = \frac{\Delta_1}{\Delta_2} = \frac{4}{7}$ Now, we want to find the ratio $BP:PC$. We know that $BC = BP + PC$. Thus, $PC = BC - BP$. $PC = BC - \frac{4}{7}BC = \frac{3}{7}BC$ Therefore, the ratio $BP:PC$ is given by: $\frac{BP}{PC} = \frac{\frac{4}{7}BC}{\frac{3}{7}BC} = \frac{4}{3}$

Step 2: Finding the Coordinates of Point P

Point $P$ divides the line segment $BC$ internally in the ratio $4:3$, where $B(-4, 3)$ and $C(-2, -5)$. Using the section formula, we can find the coordinates of $P(x_P, y_P)$: $x_P = \frac{4(-2) + 3(-4)}{4+3} = \frac{-8 - 12}{7} = \frac{-20}{7}$ $y_P = \frac{4(-5) + 3(3)}{4+3} = \frac{-20 + 9}{7} = \frac{-11}{7}$ Thus, the coordinates of point $P$ are $\left(-\frac{20}{7}, -\frac{11}{7}\right)$.

Step 3: Finding the Equation of Line AP

Line $AP$ passes through points $A(1, 1)$ and $P\left(-\frac{20}{7}, -\frac{11}{7}\right)$. The slope of $AP$ is: $m_{AP} = \frac{-\frac{11}{7} - 1}{-\frac{20}{7} - 1} = \frac{-\frac{18}{7}}{-\frac{27}{7}} = \frac{18}{27} = \frac{2}{3}$ Using the point-slope form of a line with point $A(1, 1)$: $y - 1 = \frac{2}{3}(x - 1)$ $3(y - 1) = 2(x - 1)$ $3y - 3 = 2x - 2$ $2x - 3y + 1 = 0$

Step 4: Finding the Equation of Line AC

Line $AC$ passes through points $A(1, 1)$ and $C(-2, -5)$. The slope of $AC$ is: $m_{AC} = \frac{-5 - 1}{-2 - 1} = \frac{-6}{-3} = 2$ Using the point-slope form of a line with point $A(1, 1)$: $y - 1 = 2(x - 1)$ $y - 1 = 2x - 2$ $2x - y - 1 = 0$

Step 5: Finding the X-intercepts of Lines AP and AC

To find the $x$-intercepts, we set $y = 0$ in the equations of the lines.

• Line AP: $2x - 3(0) + 1 = 0 \Rightarrow 2x = -1 \Rightarrow x = -\frac{1}{2}$. So, $X_P = \left(-\frac{1}{2}, 0\right)$.
• Line AC: $2x - (0) - 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}$. So, $X_C = \left(\frac{1}{2}, 0\right)$.

Step 6: Calculating the Area of the Triangle

The vertices of the triangle formed by lines $AP$, $AC$, and the $x$-axis are $A(1, 1)$, $X_P\left(-\frac{1}{2}, 0\right)$, and $X_C\left(\frac{1}{2}, 0\right)$. The base of the triangle is the distance between $X_P$ and $X_C$, which is: $\text{Base} = \left|\frac{1}{2} - \left(-\frac{1}{2}\right)\right| = \left|\frac{1}{2} + \frac{1}{2}\right| = 1$ The height of the triangle is the $y$-coordinate of point $A$, which is $1$. The area of the triangle is: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$

Common Mistakes & Tips

• Section Formula Mix-Up: Double-check that you are using the section formula correctly, especially when substituting the values of $m$, $n$, $x_1$, $y_1$, $x_2$, and $y_2$. A common mistake is to swap $m$ and $n$.
• Sign Errors: Be careful with signs when calculating slopes and using the point-slope form of a line.
• Ratio Confusion: Understand the difference between BP/BC and BP/PC.

Summary

We used the area ratio property to find the ratio in which point $P$ divides the line segment $BC$. Then, we used the section formula to find the coordinates of point $P$. Next, we found the equations of lines $AP$ and $AC$. Finally, we determined the $x$-intercepts of these lines and calculated the area of the triangle formed by these lines and the $x$-axis. The area enclosed by the lines $AP, AC$ and the $x$-axis is $\frac{1}{2}$.

The final answer is \boxed{\frac{1}{2}}, which corresponds to option (C).