Problem 8 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

$$ \text { For Problems } \text { find the equation of the line with the given characteristics. } $$$$ \text { Passing through point }(\sqrt{3}, \sqrt{2}) \text { and parallel to } 3 x-4 y=7 $$

Step-by-Step Solution

Answer

The equation of the line parallel to \(3x - 4y = 7\) and passing through the point \((\sqrt{3}, \sqrt{2})\) is \(y = \frac{3}{4}x + \sqrt{2} - \frac{3\sqrt{3}}{4}\)

Step 1: Identify Slope of Given Line

The first step is to find the slope of the line given the equation \(3x - 4y = 7\). This can be done by converting the standard form equation to slope-intercept form, \(y = mx + b\), where m is the slope and b is the y-intercept. To do this, first subtract 3x from both sides to obtain \(-4y = -3x + 7\), then divide by -4 on both sides to isolate y, giving \(y = \frac{3}{4}x - \frac{7}{4}\). So, the slope (m1) of the given line is \(\frac{3}{4}\).

Step 2: Determine the Slope of the Unknown Line

Since parallel lines have the same slope, the slope (m2) of the unknown line is also \(\frac{3}{4}\).

Step 3: Use the Point-Slope Form of a Line

The point-slope form of a line is \(y - y1 = m2(x - x1)\), where (x1,y1) is a point on the line and m2 is the slope. Substituting the given point \((\sqrt{3}, \sqrt{2})\) and the calculated slope into the equation we get: \(y - \sqrt{2} = \frac{3}{4}(x - \sqrt{3})\).

Step 4: Transform Equation to Standard Form

While the point-slope form is correct, it is often helpful to use the standard form, \(Ax + By = C\). So, distribute the \( \frac{3}{4}\) on the right hand side to obtain \(y - \sqrt{2} = \frac{3}{4}x - \frac{3\sqrt{3}}{4}\), add \(\sqrt{2}\) to both sides to isolate y, giving \(y = \frac{3}{4}x + \sqrt{2} - \frac{3\sqrt{3}}{4}\)

Key Concepts

Slope-Intercept Form

The slope-intercept form of a line is one of the most commonly used representations in algebra. It is given as \( y = mx + b \), where \( m \) is the slope of the line, and \( b \) is the y-intercept, which is the point where the line crosses the y-axis. This form is particularly useful because it allows you to quickly identify both the slope and the y-intercept just by looking at the equation.

To convert a standard form equation to slope-intercept form, you must solve for \( y \) to get \( y \) by itself on one side of the equation. This often involves moving terms over the equal sign and dividing by the coefficient in front of \( y \). For instance, in the given problem, the original equation \( 3x - 4y = 7 \) can be rearranged to find the slope-intercept form, resulting in \( y = \frac{3}{4}x - \frac{7}{4} \). The ease of identifying the slope \( \frac{3}{4} \) in this form highlights its usefulness.

Point-Slope Form

The point-slope form is essential when you know a point on the line and its slope. This form is written as \( y - y1 = m(x - x1) \), where \( m \) is the slope, and \( (x1, y1) \) is the known point on the line. This form is exceptionally direct for creating the equation for a line when you don't know the y-intercept.

In the exercise, after identifying the slope of the parallel line as \( \frac{3}{4} \), the point-slope form was used with the given point \( (\backslash sqrt{3}, \backslash sqrt{2}) \) to write the equation of the new line. This practically illustrates how point-slope form bridges the gap between having a point and needing the full equation of the line.

Parallel Lines

Two lines are parallel if they have the same slope and never intersect. Knowing that a line is parallel to another line immediately tells you its slope. This concept is integral when solving geometry and algebra problems.

The student exercise provided is a perfect example; it requires us to find a line parallel to \(3x - 4y = 7\). Since parallel lines have identical slopes, the slope of the new line is the same as the given one, \( \frac{3}{4} \). This pivotal attribute of parallel lines simplifies constructing equations for new lines in geometric contexts.

Standard Form Equation

The standard form of a line's equation is \( Ax + By = C \), with \( A \), \( B \), and \( C \) being integers, and \( A \) should be non-negative. This form is preferable in some algebraic operations and when emphasizing intercepts.

Transforming the point-slope or slope-intercept form to standard form might involve clearing fractions and rearranging terms, ensuring that \( A \) is positive, which is often a requirement in mathematics. Ultimately, while the standard form may not provide the immediacy of slope or intercepts as the slope-intercept form does, it holds its place for analytic geometry and is exemplified by the final conversion in the given problem.