Problem 42 - Calculus with Concepts in Calculus

Question

Find an equation of the line that is parallel to the given line \(l\) and passes through the given point \(P\). \(l: 5 x-2 y-1=0 ; P=(3,3)\)

Step-by-Step Solution

Answer

The equation is \(y = \frac{5}{2}x - \frac{9}{2}\).

Step 1: Find the slope of the given line

First, we need to find the slope of the line represented by the equation \(5x - 2y - 1 = 0\). Rewrite the equation in slope-intercept form (\(y = mx + b\)), where \(m\) is the slope. Begin by isolating \(y\):\[5x - 2y - 1 = 0 \-2y = -5x + 1 \y = \frac{5}{2}x - \frac{1}{2}\]So, the slope of the given line is \(m = \frac{5}{2}\).

Step 2: Use the point-slope form for the parallel line

Lines that are parallel share the same slope. Therefore, the line through point \(P\) that is parallel to the given line will also have a slope of \(\frac{5}{2}\). Use the point-slope form \(y - y_1 = m(x - x_1)\) where \((x_1, y_1)\) is the point \((3, 3)\):\[y - 3 = \frac{5}{2}(x - 3)\]

Step 3: Simplify the equation

Now, simplify the equation obtained in Step 2. Start by distributing \(\frac{5}{2}\):\[y - 3 = \frac{5}{2}x - \frac{15}{2}\]Add 3 to both sides to solve for \(y\):\[y = \frac{5}{2}x - \frac{15}{2} + \frac{6}{2}\]\[y = \frac{5}{2}x - \frac{9}{2}\]

Step 4: Write the final equation

The equation of the line that is parallel to the given line and passes through the point \((3,3)\) is \(y = \frac{5}{2}x - \frac{9}{2}\).

Key Concepts

Slope-Intercept Form

The slope-intercept form of a linear equation is a very popular and easy way to describe a linear equation. This form is written as \( y = mx + b \), where \( m \) represents the slope of the line and \( b \) is the y-intercept, the point where the line crosses the y-axis. The simplicity of this form lies in its ability to quickly convey the slope and the y-intercept, making it straightforward to graph the line.

To find the slope of a given line, such as \(5x - 2y - 1 = 0\), we need to rearrange it to fit into the slope-intercept form.

• First, solve for \( y \) to get \( y = mx + b \).
• Isolate \( y \) by moving the terms around and simplifying.

By doing this, you find the slope \( m \), and the resulting formula shows the linear relationship between \( x \) and \( y \). This approach helps in understanding both the direction and steepness of the line.

Point-Slope Form

The point-slope form is another very important method of representing a linear equation. It centers on using the known slope of the line \( m \) and a specific point on the line \((x_1, y_1)\) to construct the equation. This is particularly useful when you have these two key pieces of information but need to find the line's equation. The general form is \( y - y_1 = m(x - x_1) \).

This form comes in handy when you want to find the equation of a line that is parallel to another line and passes through a specific point as in the exercise. For parallel lines, the slopes are equal. So if a line through point \( P = (3,3) \) must be parallel to a given line \( y = \frac{5}{2}x - \frac{1}{2} \), it will share the same slope of \( \frac{5}{2} \).

• Substitute the point \( (3,3) \) and the slope \( \frac{5}{2} \) into \( y - y_1 = m(x - x_1) \).
• This will yield the equation of the desired line.

This form simplifies the process considerably by allowing the substitution of specific numerical values. It also helps illustrate how a change in \( x \) affects \( y \) when you know a particular point on the line.

Equation of a Line

Understanding the equation of a line is crucial in algebra. It provides a way to describe all the points that lie on that line. There are different ways to represent the equation of a line, some of which include the slope-intercept form and the point-slope form, as previously discussed.

The equation of a line encompasses the relationship between the \( x \) and \( y \) coordinates of each point along the line. With the equation \( y = \frac{5}{2}x - \frac{9}{2} \), every pair \((x, y)\) that satisfies this equation will lay on the line. The parameters, notably the slope, which controls the angle and direction of the line, and the y-intercept, which ensures its position on the graph in relation to the y-axis, are essential for accurately plotting the line.

• This form is vital for graphing and solving more complex algebraic problems.
• It allows predictions and calculations of values that lie on the line, given any one of the coordinates.
• These equations are foundational in understanding geometry and algebraic functions.

By mastering this concept, tackling problems related to lines and their graphs becomes much easier. Whether it involves finding a parallel line or interpreting complex graphs, having a solid understanding of line equations is critical.