91Ó°ÊÓ

When studying analytic geometry, determining the distance from a point to a line is a fundamental skill. This is important for various applications including determining how close two geometrical objects are. To find this distance, you use the perpendicular distance formula:

• The line equation is defined as \( ax + by + c = 0 \).
• For a point \((x_0, y_0)\), the distance \( d \) to the line is given by:\[d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}.\]

This formula measures the perpendicular or shortest distance between the point and the line. In our specific problem, the point \((0, c)\) is evaluated against two separate lines, \( ax + y = 0 \) and \( x + by = 0 \). We effectively applied this formula to calculate the individual distances using relevant coefficients from each line equation. This ensures that the concept is consistent and can be extended to other points and lines.
By understanding and applying this formula, you can tackle a wide range of geometry problems that involve distances between points and lines.

Being able to calculate the product of distances not only sharpens your math skills but also allows for comprehensive understanding in multi-faceted geometry questions.
First, you find the individual distances. For our example:

• Distance from \((0, c)\) to the line \( ax + y = 0 \) simplifies to \( d_1 = \frac{|c|}{\sqrt{a^2 + 1}} \).
• Distance from \((0, c)\) to \( x + by = 0 \) becomes \( d_2 = \frac{|bc|}{\sqrt{1 + b^2}} \).

To find the product, multiply these distances:\[d_1 \times d_2 = \frac{|c|}{\sqrt{a^2 + 1}} \times \frac{|bc|}{\sqrt{1 + b^2}} = \frac{|b c^2|}{\sqrt{(a^2 + 1)(1 + b^2)}}.\]This highlights the relationship and dependence between the distances, showcasing an important geometrical property. Understanding this concept helps in more complex scenarios where relationships between distances influence further geometric calculations.

Mathematical simplification is crucial not just to make expressions easier to understand but also to solve problems efficiently. In our example, we simplified a complex algebraic expression in the denominator:

• We started with \((a^2 + 1)(1 + b^2)\), which expanded into \(a^2(1 + b^2) + (1 + b^2)\).
• This yields \( a^2 + a^2b^2 + 1 + b^2\).
• Finally, the simplified form is \( \sqrt{a^2 + a^2b^2 + b^2 + 1} \).

Simplifying allows us to clearly see underlying relationships that could otherwise be obscured in a more complex form. It also makes the math tractable in solving further problems.
This practice not only aids in efficient problem-solving but also ensures precise mathematical communication and understanding. Mastering simplification techniques is instrumental in enhancing your overall comfort and aptitude with mathematical expressions and equations.