Problem 10 - 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 \((-3,5)\) with slope \(1 / 2\)

Step-by-Step Solution

Answer

The standard form is \(x - 2y = -13\).

Step 1: Understand the Line Equation

The point-slope form of a line's equation is given by: \[ y - y_1 = m(x - x_1) \]where \((x_1, y_1)\) is a point on the line and \(m\) is the slope.

Step 2: Substitute Given Values

Substitute the given point \((-3, 5)\) and slope \(\frac{1}{2}\) into the point-slope formula:\[ y - 5 = \frac{1}{2}(x + 3) \].

Step 3: Simplify the Equation

Distribute the slope \(\frac{1}{2}\) through the expression on the right side:\[ y - 5 = \frac{1}{2}x + \frac{3}{2} \].

Step 4: Rearrange to Standard Form

To convert to standard form \(Ax + By = C\), subtract \(\frac{1}{2}x\) and 5 from both sides:\[ -\frac{1}{2}x + y = \frac{3}{2} + 5 \].Simplify the right side:\[ -\frac{1}{2}x + y = \frac{13}{2} \].

Step 5: Clear Fractions

Multiply every term by 2 to eliminate the fraction:\[ -x + 2y = 13 \].This is now in standard form \(Ax + By = C\).

Step 6: Ensure Standard Form Conditions

Standard form typically requires that \(A\) be positive. Multiply the entire equation by -1:\[ x - 2y = -13 \].Now, the equation is in the proper standard form.

Key Concepts

Point-Slope Form

The point-slope form is a great starting point to find the equation of a line. It is useful when you have a single point and the slope of the line. The general equation for the point-slope form is:

• \( y - y_1 = m(x - x_1) \)

Here, \((x_1, y_1)\) is a known point on the line, and \(m\) represents the slope. In this context, slope refers to the rate at which the line rises or falls as you move along it.
In the exercise, the point

• \((-3,5)\)

is where the line crosses, and the slope is

• \(\frac{1}{2}\)

This tells us the line rises half a unit for each full unit it moves to the right. By plugging these values into the point-slope equation, we can start to define the line's path. This form is especially handy because it quickly incorporates the provided information—making it a great initial step when writing the equation of a line.

Standard Form

Standard form offers a tidy way to present the equation of a line by organizing it into:

• \(Ax + By = C\)

Here, \(A\), \(B\), and \(C\) are integers, and typically \(A\) should be positive. This format is valued for its neat, compact representation. It also allows for easy comparison between different line equations.
To convert from point-slope form to standard form, as in the exercise, one should rearrange terms to get:

• \(-x + 2y = 13\)

Finally, to abide by the convention of having \(A\) positive, multiply through by -1, resulting in:

• \(x - 2y = -13\)

This process shows how versatile the point-slope form is, transforming it entirely into the standard form of a line equation for clearer interpretation.

Slope-Intercept Form

The slope-intercept form is another straightforward way to express the equation of a line. It highlights both the slope and the y-intercept clearly. The format is:

• \(y = mx + b\)

Where \(m\) is the slope and \(b\) is the y-intercept (the point where the line crosses the y-axis).
Although the exercise converts to standard form, understanding slope-intercept can be helpful in many contexts. After using the point-slope form for our line:

• \(y - 5 = \frac{1}{2}x + \frac{3}{2}\)

By simplifying and rearranging:

• \(y = \frac{1}{2}x + \frac{13}{2}\)

This reveals both the rise over run aspect of the slope with its \(\frac{1}{2}\) ratio and indicates where the line meets the y-axis. This form makes graphing simple, as knowing the y-intercept gives you an easy start.