Problem 4
Question
The general equation of all non-horizontal lines in xy-plane is \(a x+b y=1\), where \(a \neq 0\)
Step-by-Step Solution
Verified Answer
The equation represents a line with slope \(-\frac{a}{b}\) and y-intercept \(\frac{1}{b}\).
1Step 1: Understanding the Equation
The given equation is \(ax + by = 1\), where \(a eq 0\). This represents a non-horizontal line because \(a\) is not zero, implying that the line has a slope and is not parallel to the x-axis.
2Step 2: Solve for y
Rearrange the general equation \(ax + by = 1\) to solve for \(y\). We get: \[ by = 1 - ax \] Now, divide every term by \(b\) to isolate \(y\): \[ y = \frac{1}{b} - \frac{a}{b}x \]
3Step 3: Identify the Slope and Y-intercept
The equation \(y = -\frac{a}{b}x + \frac{1}{b}\) is now in the slope-intercept form \(y = mx + c\), where \(m = -\frac{a}{b}\) and \(c = \frac{1}{b}\). Thus, the slope of the line is \(-\frac{a}{b}\) and the y-intercept is \(\frac{1}{b}\).
Key Concepts
Slope-Intercept FormNon-Horizontal LinesSlope and Intercept Calculation
Slope-Intercept Form
When discussing equations of lines, one of the most significant forms is the slope-intercept form. This form is represented as \(y = mx + c\), where \(m\) is the slope of the line and \(c\) is the y-intercept, or the point where the line crosses the y-axis.
This form is particularly useful because it immediately provides information about the direction and steepness of the line, as well as its intersection with the vertical axis.
This form is particularly useful because it immediately provides information about the direction and steepness of the line, as well as its intersection with the vertical axis.
- **Slope (\(m\)):** Indicates how steep the line is. A positive slope means the line goes upward as it moves from left to right, while a negative slope goes downward.
- **Y-intercept (\(c\)):** This is the value of \(y\) when \(x = 0\), showing where the line crosses the y-axis.
Non-Horizontal Lines
A non-horizontal line in the xy-plane has a noticeable characteristic: it is inclined either upwards or downwards instead of lying flat. Such lines are not parallel to the x-axis, meaning they have a slope other than zero.
Any line defined by the equation \(ax + by = 1\) where \(a eq 0\), cannot be horizontal because the presence of \(x\) with a non-zero coefficient \(a\) ensures that the line inclines. The slope equation \(m = -\frac{a}{b}\) (derived from rearranging and obtaining \(y = mx + c\)) shows that as long as \(a\) is not zero, the line has a specific degree of inclination due to the slope value \(-\frac{a}{b}\).
Without \(a\), we'd have only a vertical line defined by \(b\), which is an entirely different orientation in the plane. Thus, while horizontal lines require different considerations, most of the time you'll encounter these non-horizontal varieties in linear equations.
Any line defined by the equation \(ax + by = 1\) where \(a eq 0\), cannot be horizontal because the presence of \(x\) with a non-zero coefficient \(a\) ensures that the line inclines. The slope equation \(m = -\frac{a}{b}\) (derived from rearranging and obtaining \(y = mx + c\)) shows that as long as \(a\) is not zero, the line has a specific degree of inclination due to the slope value \(-\frac{a}{b}\).
Without \(a\), we'd have only a vertical line defined by \(b\), which is an entirely different orientation in the plane. Thus, while horizontal lines require different considerations, most of the time you'll encounter these non-horizontal varieties in linear equations.
Slope and Intercept Calculation
Calculating the slope and intercept of a line is a fundamental skill in algebra. With the equation \(y = -\frac{a}{b}x + \frac{1}{b}\) obtained from rearranging the equation \(ax + by = 1\), we identify the slope \(m\) and y-intercept \(c\) directly.
- **Slope (\(m\)):** This is calculated as \(-\frac{a}{b}\). It reflects how much the \(y\)-value changes for a unit change in \(x\). A negative slope implies the line descends as \(x\) increases.
- **Y-intercept (\(c\)):** The y-intercept here is \(\frac{1}{b}\). It tells us where the line meets the y-axis, which is when \(x = 0\).
Other exercises in this chapter
Problem 3
Equation of family of circles is \(\left(x^{2}+y^{2}-2 x-2 y+1\right)+\) $$ \lambda(3 x+4 y-7)=0 $$
View solution Problem 12
\(\frac{1}{y+1} d y=-\frac{\cos x}{2+\sin x} d x\)
View solution Problem 14
\((y \cos y+\sin y) d y=(2 x \log x+x) d x\)
View solution Problem 15
\(x^{2}(y+1) d x+y^{2}(x-1) d y=0\)
View solution