Key Concepts and Formulas

• 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 $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
• Equation of a Line (Intercept Form): The equation of a line with x-intercept $a$ and y-intercept $b$ is given by $\frac{x}{a} + \frac{y}{b} = 1$.
• Intersection of Lines: The intersection point of two lines is the point $(x, y)$ that satisfies both line equations simultaneously.

Step-by-Step Solution

1. Set up the Coordinate System

• Why this step: To translate the geometric problem into an algebraic one, enabling us to use equations of lines.
• Let the horizontal ground line AC be the x-axis.
• Let point A be the origin $(0,0)$.
• Since the pole AB is vertical and has height 15 m, point B will be at $(0, 15)$.
• Let the horizontal distance between the poles A and C be $a$ meters. So, point C will be at $(a, 0)$.
• Since the pole CD is vertical and has height 10 m, point D will be at $(a, 10)$.

2. Find the Equation of Line AD

• Why this step: Point P lies on the line segment AD. We need its equation to represent all points on it.
• Line AD passes through points A $(0,0)$ and D $(a,10)$.
• Using the two-point form of a line, $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$: $y - 0 = \frac{10 - 0}{a - 0}(x - 0)$ $y = \frac{10}{a}x \quad \text{(Equation 1)}$

3. Find the Equation of Line BC

• Why this step: Point P also lies on the line segment BC. We need its equation as well.
• Line BC passes through points B $(0,15)$ and C $(a,0)$.
• Using the intercept form of a line, $\frac{x}{x_{\text{intercept}}} + \frac{y}{y_{\text{intercept}}} = 1$: The x-intercept is $a$ (point C). The y-intercept is $15$ (point B). $\frac{x}{a} + \frac{y}{15} = 1 \quad \text{(Equation 2)}$

4. Find the Intersection Point P

• Why this step: Point P is the intersection of lines AD and BC. Its coordinates must satisfy both Equation 1 and Equation 2 simultaneously. We need to solve this system of equations for $x$ and $y$. The $y$-coordinate of P will be its height.
• From Equation 1, we can express $x$ in terms of $y$: $x = \frac{ay}{10}$
• Substitute this expression for $x$ into Equation 2: $\frac{\left(\frac{ay}{10}\right)}{a} + \frac{y}{15} = 1$
• Simplify the first term: $\frac{y}{10} + \frac{y}{15} = 1$
• To combine the terms on the left side, find a common denominator, which is 30 (LCM of 10 and 15): $\frac{3y}{30} + \frac{2y}{30} = 1$
• Combine the numerators: $\frac{3y + 2y}{30} = 1$ $\frac{5y}{30} = 1$
• Simplify the fraction: $\frac{y}{6} = 1$
• Solve for $y$: $y = 6$

5. Determine the Height of P

• Why this step: The y-coordinate of any point in this setup represents its vertical distance from the x-axis (which is the ground line AC).
• The y-coordinate of point P is 6.
• Therefore, the height of point P above the line AC is 6 meters.

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful with algebraic manipulations, especially when dealing with fractions. Double-check your work at each step.
• Alternative Methods: Consider alternative solution approaches, such as similar triangles, as they may provide a more efficient solution path.
• Coordinate System Placement: A well-chosen coordinate system can significantly simplify the problem.

Summary

By establishing a coordinate system and determining the equations of the lines connecting the tops of the poles to the bases of the opposing poles, we located their intersection point. The y-coordinate of this intersection point directly corresponds to the height of P above the ground, which is 6 meters.

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