Problem 33 - 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 \((5,-1)\) and perpendicular to the line passing through \((-2,1)\) and \((1,-2)\)

Step-by-Step Solution

Answer

The standard form is \(x - y - 6 = 0\).

Step 1: Find the slope of the given line

The slope of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is calculated using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For the points \((-2,1)\) and \(1,-2)\), the slope \(m\) is \(\frac{-2 - 1}{1 - (-2)} = \frac{-3}{3} = -1\).

Step 2: Find the perpendicular slope

Two lines are perpendicular if the product of their slopes is -1. The slope of a line perpendicular to one with slope \(m\) is given by \(-\frac{1}{m}\). Since the slope of the given line is \(-1\), the perpendicular slope is \(1\).

Step 3: Use point-slope form with the new slope

Using the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line. We apply this to the point \((5, -1)\) with slope \(1\), resulting in \(y + 1 = 1(x - 5)\).

Step 4: Simplify and convert to standard form

Expand the equation from step 3: \(y + 1 = x - 5\). Rearrange it to get terms on one side: \(x - y - 6 = 0\). This is the line equation in standard form, which is generally written as \(Ax + By + C = 0\).

Key Concepts

Slope of a Line

When we talk about the slope of a line, we are essentially measuring its steepness. The slope is commonly denoted by the symbol \(m\). For any line, the slope can be calculated using the formula:\[m = \frac{y_2 - y_1}{x_2 - x_1}\]
This formula finds the change in the y-coordinates \(y_2 - y_1\) divided by the change in the x-coordinates \(x_2 - x_1\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) on the line. Simply put, it tells us how much the line rises or falls as we move from one point to another along the x-axis.
For example, in the original exercise, the slope was calculated for the points \((-2, 1)\) and \((1, -2)\). With these points, we found that:

• Change in y: \(-2 - 1 = -3\)
• Change in x: \(1 - (-2) = 3\)
• Slope \(m = \frac{-3}{3} = -1\)

This way, we know this line is decreasing because the slope is negative.

Point-Slope Form

The point-slope form is particularly useful when we know a point on the line and the slope. The formula 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.
This form is especially handy for quickly writing the equation of a line when you have only one point and the slope available.
In our exercise, we used the point-slope form to find the equation of a line perpendicular to the given line and passing through a known point \((5, -1)\) with perpendicular slope \(1\). Thus, our specific equation began as:
\[y + 1 = 1(x - 5)\]
This equation uses the known point and the new perpendicular slope to describe the new line.

Standard Form of a Line

Lines are often represented in what is known as the standard form. The standard form is expressed as:
\[Ax + By + C = 0\]
where \(A\), \(B\), and \(C\) are integers, and \(A\) should be non-negative. This form is convenient for certain applications, such as finding intersections and for solving systems of equations.
To transform an equation from point-slope or slope-intercept form to standard form, you'll typically need to move terms around to arrange them as shown above. In our exercise, starting from the point-slope form \(y + 1 = x - 5\), we rearranged to have:
\[x - y - 6 = 0\]
This rearrangement is performed by combining like terms and ensuring all variables are on one side of the equation. The result is a cleaner, standardized format that is often preferred.