Finding The Equation Of A Line Perpendicular To BC Passing Through Point A

In the realm of coordinate geometry, a fundamental problem involves finding the equation of a line that satisfies specific conditions. This article delves into the process of determining the equation of a line that passes through a given point and is perpendicular to a line segment defined by two other points. We will explore this concept through the specific example of triangle $ABC$, where we aim to find the equation of a line passing through point $A$ and perpendicular to side $\overline{BC}$.

Problem Statement

Consider triangle $ABC$ with vertices defined by the points $A(3,8)$, $B(7,5)$, and $C(2,3)$. Our objective is to find the equation of a line that passes through point $A$ and is perpendicular to the line segment $\overline{BC}$. This problem combines concepts of coordinate geometry, including slope, perpendicular lines, and the point-slope form of a linear equation. Understanding these concepts is crucial for solving this problem effectively.

Step-by-Step Solution

1. Calculate the Slope of BC

To begin, we need to determine the slope of the line segment $\overline{BC}$. The slope, denoted as $m$, is a measure of the steepness of a line and is calculated as the change in the y-coordinate divided by the change in the x-coordinate. The formula for the slope between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Applying this formula to points $B(7,5)$ and $C(2,3)$, we have:

$m_{BC} = \frac{3 - 5}{2 - 7} = \frac{-2}{-5} = \frac{2}{5}$

Thus, the slope of $\overline{BC}$ is $\frac{2}{5}$. This value is essential as it will help us determine the slope of the perpendicular line.

2. Determine the Slope of the Perpendicular Line

Now, we need to find the slope of the line that is perpendicular to $\overline{BC}$. A fundamental property of perpendicular lines is that their slopes are negative reciprocals of each other. This means that if the slope of one line is $m$, the slope of a line perpendicular to it is $-\frac{1}{m}$.

Since the slope of $\overline{BC}$ is $\frac{2}{5}$, the slope of the line perpendicular to it, denoted as $m_{\perp}$, is:

$m_{\perp} = -\frac{1}{\frac{2}{5}} = -\frac{5}{2}$

Therefore, the slope of the line we are seeking is $-\frac{5}{2}$. This negative reciprocal relationship is a cornerstone of coordinate geometry, ensuring that the lines intersect at a right angle.

3. Use the Point-Slope Form to Find the Equation

The point-slope form of a linear equation is a powerful tool for finding the equation of a line when we know its slope and a point it passes through. The point-slope form is given by:

$y - y_1 = m(x - x_1)$

where $(x_1, y_1)$ is a point on the line and $m$ is the slope. We know that the line passes through point $A(3,8)$ and has a slope of $-\frac{5}{2}$. Substituting these values into the point-slope form, we get:

$y - 8 = -\frac{5}{2}(x - 3)$

This equation represents the line in point-slope form. However, it is often preferable to express the equation in slope-intercept form, which we will do in the next step.

4. Convert to Slope-Intercept Form

The slope-intercept form of a linear equation is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. To convert the equation from point-slope form to slope-intercept form, we need to solve for $y$. Starting with the equation:

$y - 8 = -\frac{5}{2}(x - 3)$

First, distribute the slope $-\frac{5}{2}$ across the terms inside the parentheses:

$y - 8 = -\frac{5}{2}x + \frac{15}{2}$

Next, add 8 to both sides of the equation to isolate $y$:

$y = -\frac{5}{2}x + \frac{15}{2} + 8$

To combine the constants, we need to express 8 as a fraction with a denominator of 2:

$8 = \frac{16}{2}$

So, the equation becomes:

$y = -\frac{5}{2}x + \frac{15}{2} + \frac{16}{2}$

Now, combine the fractions:

$y = -\frac{5}{2}x + \frac{31}{2}$

Thus, the equation of the line passing through point $A$ and perpendicular to $\overline{BC}$ in slope-intercept form is:

$y = -\frac{5}{2}x + \frac{31}{2}$

Conclusion

In conclusion, we have successfully determined the equation of a line that passes through point $A(3,8)$ and is perpendicular to the line segment $\overline{BC}$ in triangle $ABC$. The step-by-step solution involved calculating the slope of $\overline{BC}$, finding the negative reciprocal of this slope to determine the slope of the perpendicular line, using the point-slope form of a linear equation, and converting the equation to slope-intercept form. This comprehensive approach demonstrates the application of key concepts in coordinate geometry and provides a clear methodology for solving similar problems. The final equation, $y = -\frac{5}{2}x + \frac{31}{2}$, accurately represents the line that meets the specified conditions.

This process underscores the importance of understanding the relationships between slopes of perpendicular lines and the various forms of linear equations. By mastering these concepts, one can effectively tackle a wide range of geometric problems involving lines and points in the coordinate plane. The ability to translate geometric conditions into algebraic equations is a fundamental skill in mathematics and is crucial for problem-solving in various fields.

By working through this problem, we have not only found a specific solution but also reinforced our understanding of the underlying principles of coordinate geometry. This knowledge can be applied to more complex problems and serves as a solid foundation for further study in mathematics and related disciplines. The methodical approach demonstrated here, from calculating slopes to using the point-slope form and converting to slope-intercept form, is a valuable technique that can be adapted to many similar scenarios.

Therefore, the equation $y = -\frac{5}{2}x + \frac{31}{2}$ is the definitive solution to the problem, providing a clear and concise representation of the line that satisfies the given conditions. This example highlights the power of coordinate geometry in bridging the gap between geometric concepts and algebraic expressions, allowing us to solve complex problems in a systematic and logical manner.