Problem 34 - Calculus for Biology and Medicine

Question

Determine the equation of the line that satisfies the stated requirements. Put the equation in standard form. The line passing through \((4,-1)\) and perpendicular to the line passing through \((-2,0)\) and \((1,1)\)

Step-by-Step Solution

Answer

The equation of the line is \( 3x + y = 11 \) in standard form.

Step 1: Find the slope of the given line

Calculate the slope \( m_1 \) of the line passing through the points \((-2,0)\) and \((1,1)\) using the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Substitute the given points: \( m_1 = \frac{1 - 0}{1 - (-2)} = \frac{1}{3} \). So, the slope of the given line is \( \frac{1}{3} \).

Step 2: Determine the perpendicular slope

A line that is perpendicular to another has a slope that is the negative reciprocal. Therefore, the slope \( m_2 \) of the line we want to find is \(-3\). This is because \(-3 \times \frac{1}{3} = -1\).

Step 3: Write the point-slope form

Using the point-slope form formula \( y - y_1 = m(x - x_1) \), where \((x_1, y_1) = (4,-1)\) and \( m = -3 \), substitute the values: \( y + 1 = -3(x - 4) \).

Step 4: Simplify to slope-intercept form

Distribute \( -3 \) in the equation from step 3: \( y + 1 = -3x + 12 \). Simplify this to \( y = -3x + 11 \) by subtracting 1 from both sides.

Step 5: Convert to standard form

Rearrange the equation from step 4 into standard form \( Ax + By = C \) by adding \( 3x \) to both sides: \( 3x + y = 11 \). This is the equation in standard form.

Key Concepts

Understanding Slope

The slope of a line is a measure of its steepness and direction. It's a crucial concept when dealing with linear equations. To find the slope of a line, you need two points, often labeled as \((x_1, y_1)\) and \((x_2, y_2)\). The formula for calculating the slope \(m\) is:

• \( m = \frac{y_2 - y_1}{x_2 - x_1} \)

This formula provides the rate of change between the two points. In this exercise, we used the points \((-2,0)\) and \((1,1)\) to find a slope of \(\frac{1}{3}\).
An important aspect of slopes is understanding perpendicular lines. A line perpendicular to another will have a slope that is the negative reciprocal of the original line's slope. Given our initial slope \(\frac{1}{3}\), the perpendicular slope becomes \(-3\) because \(-3 \times \frac{1}{3} = -1\), demonstrating their perpendicular nature.

Using the Point-Slope Form

The point-slope form of a linear equation is useful for writing the equation of a line when you know its slope and a point on the line. It takes the form:

• \(y - y_1 = m(x - x_1)\)

Here, \((x_1, y_1)\) is any point on the line, and \(m\) is the slope. For the line we are dealing with, passing through the point \((4, -1)\) and having a slope of \(-3\), the equation becomes:
\(y + 1 = -3(x - 4)\).
This form makes it easy to start building the equation of the line. You simply plug in the values for the slope and the chosen point. From here, you can easily transform the equation into other forms like the slope-intercept form.

Transforming to Standard Form of a Line

The standard form of a line is an organized way of writing a linear equation. It is expressed as:

• \(Ax + By = C\)

Where \(A\), \(B\), and \(C\) are integers, and \(A\) should be non-negative. From the point-slope form \(y + 1 = -3(x - 4)\), we first distribute and rearrange into a slope-intercept form \(y = -3x + 11\).
The next step is transforming this into standard form. We move \(-3x\) over to the same side as \(y\), yielding:

• \(3x + y = 11\)

This is our desired line equation in standard form, neatly organizing the terms.