Problem 38 - Precalculus: Functions and Graphs

Question

Find equations for the altitudes of the triangle with vertices \(A(-3,2), B(5,4),\) and \(C(3,-8),\) and find the point at which the altitudes intersect.

Step-by-Step Solution

Answer

The altitudes intersect at the point \(\left(\frac{15}{23}, \frac{7}{23}\right)\).

Step 1: Find Slope of Side BC

Calculate the slope of the line segment connecting points B(5,4) and C(3,-8) using the slope formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Thus, \( m = \frac{-8 - 4}{3 - 5} = \frac{-12}{-2} = 6 \).

Step 2: Equation of Altitude from Vertex A

The altitude from vertex A is perpendicular to side BC. Since the slope of BC is 6, the slope of the altitude is \( -\frac{1}{6} \). Using point A(-3,2), use the point-slope form \( y - y_1 = m(x - x_1) \): \( y - 2 = -\frac{1}{6}(x + 3) \), which simplifies to \( y = -\frac{1}{6}x + \frac{1}{2} \).

Step 3: Find Slope of Side AC

Calculate the slope of line AC where A(-3,2) and C(3,-8): \( m = \frac{-8 - 2}{3 + 3} = \frac{-10}{6} = -\frac{5}{3} \).

Step 4: Equation of Altitude from Vertex B

The altitude from vertex B is perpendicular to side AC. The slope of the altitude is the negative reciprocal of \(-\frac{5}{3}\), which is \(\frac{3}{5}\). Using point B(5,4) in the point-slope form gives: \( y - 4 = \frac{3}{5}(x - 5) \), which simplifies to \( y = \frac{3}{5}x + 1 \).

Step 5: Find Slope of Side AB

Calculate the slope of line AB where A(-3,2) and B(5,4): \( m = \frac{4 - 2}{5 + 3} = \frac{2}{8} = \frac{1}{4} \).

Step 6: Equation of Altitude from Vertex C

The altitude from vertex C is perpendicular to side AB. The slope of the altitude is \(-4\), as it is the negative reciprocal of \(\frac{1}{4}\). Using point C(3,-8), the equation is \( y + 8 = -4(x - 3) \), which simplifies to \( y = -4x + 4 \).

Step 7: Find Intersection of Altitudes

Solve the equations from Steps 2 and 4 to find their intersection. Set \( -\frac{1}{6}x + \frac{1}{2} = \frac{3}{5}x + 1 \). Solving gives \(\frac{23}{30}x = \frac{1}{2}\), leading to \(x = \frac{15}{23}\). Substitute back to find \(y\): \( y = \frac{3}{5}\times\frac{15}{23} + 1 = \frac{7}{23} \).

Key Concepts

Equations of Lines

Understanding the concept of an equation of a line is fundamental in coordinate geometry. A line's equation symbolizes all the points that lie along it. The most common form used to write an equation of a line is the "slope-intercept form," which is expressed as \( y = mx + b \). In this expression, \( m \) is the slope of the line, and \( b \) is the y-intercept.
This form tells us two things:

• The slope \( m \) indicates how steep the line is.
• The y-intercept \( b \) is the point where the line crosses the y-axis.

Another useful form is the "point-slope form," expressed as \( y - y_1 = m(x - x_1) \), which is particularly handy when you know a point on the line \((x_1, y_1)\) and the slope \( m \), and want to find the complete equation. This is the form often used to find equations of altitudes and other lines related to given points.

Slope of a Line

The slope of a line is a measure of its steepness and the direction in which it inclines. It is calculated as the change in the y-coordinate divided by the change in the x-coordinate between two points on the line. This relationship is expressed with the formula:
\[m = \frac{y_2 - y_1}{x_2 - x_1} \]It is important to note:

• If the slope is positive, the line inclines upward from left to right.
• If the slope is negative, the line declines downward from left to right.
• If the slope is zero, the line is perfectly horizontal.
• If the slope is undefined (division by zero), the line is vertical.

Knowing the slope is crucial for solving problems involving perpendicular and parallel lines, as it helps determine the nature of the relationship between them.

Intersection of Lines

The point of intersection of two lines is where they meet or cross each other on a plane. In terms of algebra, this means finding values \(x\) and \(y\) that satisfy both equations of the lines simultaneously.
To find this intersection point:

• Set the equations equal to each other and solve for \(x\).
• Substitute the found \(x\) value back into one of the original equations to find \(y\).

For example, given two line equations \( y = -\frac{1}{6}x + \frac{1}{2} \) and \( y = \frac{3}{5}x + 1 \), set them equal to solve for \(x\):\[-\frac{1}{6}x + \frac{1}{2} = \frac{3}{5}x + 1\]Solve this to find the x-coordinate of the intersection and substitute back to find the y-coordinate. The intersection point of these lines is essential when determining a triangle's orthocenter, where its altitudes intersect.

Perpendicular Lines

Perpendicular lines intersect at a right angle (90 degrees). In coordinate geometry, when two lines are perpendicular, the slopes of these lines multiply to give \(-1\). Hence, if one line has a slope \( m_1 \), the slope \( m_2 \) of a line perpendicular to it is given by:
\[ m_2 = -\frac{1}{m_1} \]This relationship is key when determining the equations of altitudes of a triangle, as each altitude is perpendicular to a side of the triangle. To find an equation of such a line, you begin by identifying the slope of one side of the triangle, calculate its negative reciprocal, and use it as the slope for the altitude. Using this concept ensures that the constructed altitude is perfectly perpendicular to the side it relates to.