Problem 35 - Algebra and Trigonometry

Question

Find an equation of the line that satisfies the given conditions. Through \((-1,-2) :\) perpendicular to the line \(2 x+5 y+8=0\)

Step-by-Step Solution

Answer

The equation of the line is \(5x - 2y = 8\).

Step 1: Identify the Slope of the Given Line

The equation of the given line is \(2x + 5y + 8 = 0\). We will express it in the slope-intercept form \(y = mx + c\) to find its slope. Rearrange the equation:\[5y = -2x - 8\] \[y = -\frac{2}{5}x - \frac{8}{5}\]The slope \(m\) of this line is \(-\frac{2}{5}\).

Step 2: Determine the Slope of the Perpendicular Line

Lines that are perpendicular to each other have slopes that are negative reciprocals. Given the slope of the original line is \(-\frac{2}{5}\), the slope \(m_p\) of the perpendicular line is the negative reciprocal:\[m_p = \frac{5}{2}\]

Step 3: Use Point-Slope Formula for the Equation

To find the equation of the line that is perpendicular and passes through the point \((-1, -2)\), we use the point-slope form:\[y - y_1 = m(x - x_1)\]where \((x_1, y_1)\) is the point \((-1, -2)\) and \(m = \frac{5}{2}\).

Step 4: Substitute the Values into the Equation

Substitute the point \((-1, -2)\) and slope \(\frac{5}{2}\) into the point-slope form:\[y - (-2) = \frac{5}{2}(x - (-1))\]Simplify the equation:\[y + 2 = \frac{5}{2}(x + 1)\]

Step 5: Rearrange into Slope-Intercept or Standard Form

To express the equation, distribute and simplify:\[y + 2 = \frac{5}{2}x + \frac{5}{2}\]Subtract 2 from both sides:\[y = \frac{5}{2}x + \frac{5}{2} - 2\]Simplify further:\[y = \frac{5}{2}x - \frac{4}{0}\]Now, for a standard form, multiply through by 2 to remove fractions:\[2y = 5x - 8\]and rearrange:\[5x - 2y = 8\]

Key Concepts

Slope-Intercept Form

The slope-intercept form is a straightforward way to express the equation of a line. It is given by the formula \( y = mx + c \), where \( m \) is the slope of the line, and \( c \) is the y-intercept where the line crosses the y-axis.
This form is particularly useful because it immediately tells us two important things about the line:

• The slope \( m \) indicates how steep the line is. If \( m \) is positive, the line rises as it moves from left to right; if negative, it falls.
• The y-intercept \( c \) shows the point where the line intersects the y-axis.

To find the slope of a line from an equation not in this form, you need to rearrange the equation to solve for \( y \). For example, transforming an equation from the general form like \( 2x + 5y + 8 = 0 \) into the slope-intercept form involves isolating \( y \) on one side. It can simplify understanding the line's behavior and can make graphing it easier.

Point-Slope Form

When you need the equation of a line and you know a point on the line and the slope, then the point-slope form is your go-to format. This form is written as \( y - y_1 = m(x - x_1) \). Here, \((x_1, y_1)\) is the point through which the line passes, and \( m \) is the slope. This is particularly helpful when you have the specific coordinates of the point.

• Perfect for deriving an equation when details of any single point and the slope are known.
• It easily converts into slope-intercept form by rearranging the equation.

This form is flexible and can quickly lead you to slope-intercept or standard forms, making it a user-friendly tool for understanding and describing lines.

Negative Reciprocal

To find the slope of a line perpendicular to another, you need to use the concept of a negative reciprocal. Two lines are perpendicular if the product of their slopes is \(-1\). This means if the slope of the original line is \( m \), a perpendicular line's slope will be \(-1/m\).
This technique is crucial when finding equations of perpendicular lines because:

• It offers a simple mathematical method for finding the new slope needed for perpendicularity.
• For instance, if a line has a slope of \(-\frac{2}{5}\), the perpendicular slope will be \( \frac{5}{2} \).

Understanding how to calculate and apply the negative reciprocal ensures you can effectively determine perpendicular lines in geometric problems.