Problem 23 - The Calculus with Analytic Geometry

Question

Given the line \(l\), having the equation \(A x+B y+C=0, B \neq 0\), find (a) the slope, (b) the \(y\) intercept, (c) the \(x\) intercept (d) an equation of the line through the origin perpendicular to \(l\).

Step-by-Step Solution

Answer

Slope: \[ -\frac{A}{B} \] \[ y \]-intercept: \[ -\frac{C}{B} \] \[ x \]-intercept: \[ -\frac{C}{A} \]Equation of perpendicular line: \[ y = \frac{B}{A} x \]

Step 1 - Identify the standard form

The given line has the equation in the standard form: \[ Ax + By + C = 0 \].

Step 2 - Find the slope

Rearrange the equation to the slope-intercept form: \[ By = -Ax - C \] Divide by \[ B \] to isolate \[ y \]: \[ y = -\frac{A}{B}x - \frac{C}{B} \]. The slope is: \[ m = -\frac{A}{B} \].

Step 3 - Determine the y-intercept

In the slope-intercept form \[ y = -\frac{A}{B}x - \frac{C}{B} \] the \[ y \]-intercept is the constant term: \[ -\frac{C}{B} \].

Step 4 - Calculate the x-intercept

To find the x-intercept, set \[ y = 0 \] in the original equation: \[ Ax + B(0) + C = 0 \]. Solve for \[ x \]: \[ x = -\frac{C}{A} \].

Step 5 - Equation of perpendicular line through the origin

A line perpendicular to \[ l \] will have a slope that is the negative reciprocal of the slope of \[ l \]. The slope of \[ l \] is \[ -\frac{A}{B} \], so the perpendicular slope is \[ \frac{B}{A} \]. The equation of a line through the origin with this slope is: \[ y = \frac{B}{A} x \].

Key Concepts

Slope-Intercept Form (y = mx + b)

Understanding the slope-intercept form is crucial for working with linear equations. The slope-intercept form of a line is expressed as \(y = mx + b\), where \(m\) represents the slope of the line, and \(b\) represents the y-intercept.

The slope, \(m\), indicates how steep the line is, and it is calculated by the formula \(-\frac{A}{B}\) when the line is given in the standard form \(Ax + By + C = 0\).
To convert an equation from standard form to slope-intercept form:

• Isolate y by solving the equation for y.
• Identify the slope (m) as the coefficient of x.
• Identify the y-intercept (b) as the constant term.

This process helps to directly visualize the slope and the y-intercept in the graph of a line.

X-Intercept

The x-intercept is the point where a line crosses the x-axis. At this point, the value of y is zero.
To find the x-intercept of the line given by the standard form equation \(Ax + By + C = 0\):

• Set \(y = 0\) in the equation.
• Solve for x.

For example, substituting \(y = 0\): \Ax + B(0) + C = 0\ simplifies to \Ax + C = 0\. Solving for x gives the x-intercept as \(-\frac{C}{A}\).

The x-intercept is crucial for understanding where the line intersects the horizontal axis on a graph.

Y-Intercept

The y-intercept is the point where a line crosses the y-axis. At this point, the value of x is zero.
To find the y-intercept from the standard form equation \(Ax + By + C = 0\):

• Set \(x = 0\) in the equation.
• Solve for y.

When converted to slope-intercept form \(-\frac{C}{B}\), the y-intercept becomes the constant term.

Substituting \(x = 0\) gives: \By = -C\, and solving for y provides the y-intercept \(-\frac{C}{B}\).
The y-intercept allows us to identify the starting point of the line on the vertical axis in a graph.

Perpendicular Lines

Two lines are perpendicular if the product of their slopes is -1.
If you have a line with a slope \(-\frac{A}{B}\), a line perpendicular to it will have a slope that is the negative reciprocal. In this case, the perpendicular slope would be \frac{B}{A}\.

For a line through the origin (0,0) and perpendicular to the line given by \(Ax + By + C = 0\):

• Identify the perpendicular slope.
• Use the point-slope form to write the equation.

The equation of the line becomes \(y = \frac{B}{A}x\), ensuring the new line is perfectly perpendicular and passes through the origin.
This concept is essential for understanding orthogonal relationships in geometry and graphing.