Problem 35 - Precalculus

Question

Find an equation for the line that is described. Write the answer in the two forms \(y=m x+b\) and \(A x+B y+C=0\). Is perpendicular to \(4 y-3 x=1\) and passes through (4,0).

Step-by-Step Solution

Answer

The equation in \(y = mx + b\) is \(y = -\frac{4}{3}x + \frac{16}{3}\), and in \(Ax + By + C = 0\) form, it is \(4x + 3y - 16 = 0\).

Step 1: Identify the Slope of the Given Line

The given line's equation is written as \(4y - 3x = 1\). To find the slope, rewrite the equation in slope-intercept form \(y = mx + b\). Divide all terms by 4 to get \(y = \frac{3}{4}x + \frac{1}{4}\). Therefore, the slope \(m\) of this line is \(\frac{3}{4}\).

Step 2: Determine the Slope of the Perpendicular Line

The slope of a line perpendicular to another is the negative reciprocal of the original slope. Therefore, take the negative reciprocal of \(\frac{3}{4}\), giving a slope of \(-\frac{4}{3}\) for the perpendicular line.

Step 3: Use Point-Slope Form to Find the Equation

The line perpendicular to the given passes through the point (4,0). Using the point-slope form equation, \(y - y_1 = m(x - x_1)\), substitute \(m = -\frac{4}{3}\), \(x_1 = 4\), and \(y_1 = 0\). This gives the equation \(y - 0 = -\frac{4}{3}(x - 4)\).

Step 4: Simplify to Slope-Intercept Form

Expand the equation from Step 3: \(y = -\frac{4}{3}(x - 4)\). Simplify it to get \(y = -\frac{4}{3}x + \frac{16}{3}\). Thus, the equation in slope-intercept form is \(y = -\frac{4}{3}x + \frac{16}{3}\).

Step 5: Convert to General Form

From the slope-intercept form \(y = -\frac{4}{3}x + \frac{16}{3}\), multiply through by 3 to eliminate fractions: \(3y = -4x + 16\). Rearrange to general form as \(4x + 3y - 16 = 0\).

Key Concepts

Slope-Intercept Form

The slope-intercept form of a linear equation is one of the easiest ways to express a line. This form is written as \( y = mx + b \), where \( m \) indicates the slope, and \( b \) is the y-intercept, which is where the line crosses the y-axis. This form is very handy because it quickly tells us both the steepness of the line (the slope) and its starting point on the y-axis.
The process to find the equation of a line in the slope-intercept form when given a line that is perpendicular to another line involves a few steps. First, we need the slope of the original line, calculated by getting the line equation into the slope-intercept form. For example, if the original line is \( 4y - 3x = 1 \), you would divide through by 4 to isolate \( y \) and get \( y = \frac{3}{4}x + \frac{1}{4} \), revealing a slope of \( \frac{3}{4} \) for the original line.
To find the slope of a line perpendicular to this original line, take the negative reciprocal of the original slope. The negative reciprocal of \( \frac{3}{4} \) is \( -\frac{4}{3} \). Now, knowing the slope \( -\frac{4}{3} \) and a point the line goes through, for example, (4,0), you can use this information with the point-slope formula to find and simplify the new line's equation in slope-intercept form.

Point-Slope Form

The point-slope form is a valuable template for determining a line's equation when a point on the line and the slope are known. It's expressed as \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \( (x_1, y_1) \) is a known point on the line. This format is especially useful when the slope is given, and you want to transition into an equation recognizing an exact point.
For perpendicular lines, after determining the perpendicular slope from your given equation, insert it, along with the coordinates of a known point into the point-slope form. Let's say we have a line passing through (4,0) with a perpendicular slope of \( -\frac{4}{3} \). Place these variables into the point-slope expression: \( y - 0 = -\frac{4}{3}(x - 4) \). This direct application makes it immediately clear how the slope and point relate, providing a robust framework for solving the equation.
The point-slope form is particularly effective in transitioning to other forms as it serves almost like a bridge, converting specific informational aspects regarding a line into a standard mathematical expression.

General Form

The general form of a linear equation is another essential way to represent lines on a plane. It’s written as \( Ax + By + C = 0 \) where \( A \), \( B \), and \( C \) are integers, with \( A \) typically taking a positive value. This form may seem less intuitive at first, as the slope and intercepts aren't immediately visible, but it's highly beneficial for analyzing the relationship between line coefficients and when working with systems of equations.
To convert a line from the slope-intercept form to the general form, you typically start by eliminating fractions (if any) by multiplying through by the least common multiple. Once fractions are eliminated, rearrange terms to get all expressions on one side of the equation, yielding \( Ax + By + C = 0 \). For example, if we have the slope-intercept form \( y = -\frac{4}{3}x + \frac{16}{3} \), multiply the whole equation by 3 to remove the fraction, resulting in \( 3y = -4x + 16 \). Rearrange this to \( 4x + 3y - 16 = 0 \), representing the general form.
The process of rewriting equations in this manner solidifies understanding of linear algebra and helps when plotting on a graph or considering equations in a broader mathematical context.