Problem 18 - Precalculus

Question

Find an equation for the line passing through the two given points. Write your answer in the form \(y=m x+b\). (a) (7,9) and (-11,9) (b) \((5 / 4,2)\) and \((3 / 4,3)\) (c) (12,13) and (13,12)

Step-by-Step Solution

Answer

(a) \( y = 9 \); (b) \( y = -2x + 5.5 \); (c) \( y = -x + 25 \).

Step 1: Understanding Slope Formula

To find the equation of a line, we first need to calculate its slope, denoted as \( m \). The formula for slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \].

Step 2: Calculate Slope for (a)

Substituting the points (7,9) and (-11,9) into the slope formula, we have: \[ m = \frac{9 - 9}{-11 - 7} = \frac{0}{-18} = 0 \]. Since the slope is 0, the line is horizontal.

Step 3: Finding Equation for (a)

For a horizontal line, the equation is \( y = c \) where \( c \) is the constant y-value for all points on the line. Thus, for problem (a), the equation is \( y = 9 \).

Step 4: Calculate Slope for (b)

Using the slope formula for points \((5/4, 2)\) and \((3/4, 3)\): \[ m = \frac{3 - 2}{3/4 - 5/4} = \frac{1}{-2/4} = -2 \].

Step 5: Finding Equation for (b)

With \( m = -2 \), use point-slope form, \( y - y_1 = m(x - x_1) \) with \( (x_1, y_1) = (5/4, 2) \). \[ y - 2 = -2(x - 5/4) \], simplifying gives \( y = -2x + 5.5 \).

Step 6: Calculate Slope for (c)

Applying the same formula to points (12,13) and (13,12): \[ m = \frac{12 - 13}{13 - 12} = -1 \].

Step 7: Finding Equation for (c)

Using \( m = -1 \) and point-slope form with the point (12, 13), we get: \[ y - 13 = -1(x - 12) \]. Simplifying, the equation is \( y = -x + 25 \).

Key Concepts

Slope

The slope is a fundamental concept in understanding linear equations. It indicates the steepness of a line and is often denoted by the letter "m". The slope is calculated using two points on the line. To find the slope between two points, \( (x_1, y_1) \) and \( (x_2, y_2) \), use the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] This formula tells us the rate of change of the y-value with respect to the x-value.
For example, in problem (a), the points (7, 9) and (-11, 9) share the same y-coordinate. So, the slope is calculated as: \[ m = \frac{9 - 9}{-11 - 7} = \frac{0}{-18} = 0 \] This means the line is horizontal, showing that the slope can help us determine the line's orientation.
In contrast, different y-values result in non-zero slopes, like in problem (b), where the slope was \( -2 \) indicating a line falling as it moves from left to right. Understanding slope is crucial as it lays the groundwork for writing the equation of a line.

Point-Slope Form

The point-slope form is an essential tool in forming the equation of a line when you know a point on the line and its slope. This form is expressed as: \[ y - y_1 = m(x - x_1) \] where \( m \) is the slope and \( (x_1, y_1) \) is a point on the line.
Let's see it in action with problem (b). We found that the slope \( m = -2 \). Using the point (5/4, 2) in our point-slope formula gives us: \[ y - 2 = -2(x - 5/4) \] This allows us to express the line equation in a simpler form, \( y = -2x + 5.5 \), by expanding and simplifying.
Point-slope form is beneficial because it directly incorporates a specific point and the slope to build the equation. This can make it faster than other methods in certain situations and helps visualize how the line behaves.

Horizontal Line Equation

A horizontal line has a characteristic feature where its y-values remain constant irrespective of the x-values. This results in a slope of zero. In practical terms, the equation for a horizontal line is given by \( y = c \), where \( c \) is the constant y-value for all points on the line.
In problem (a), where the points were (7, 9) and (-11, 9), we observed that the y-values were constant at 9. Since the slope calculated as 0, it confirmed that the line was horizontal. Therefore, the equation is \( y = 9 \).
Understanding horizontal lines is important because they represent situations where there is no vertical change as you move through the x-axis. They tell us about stability and uniformity in data, and recognizing them helps quickly identify particular line characteristics.