Problem 45 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 45-48, find the coordinates of the point of intersection. Then write an equation for the line through that point perpendicular to the line given first. \(2 x+3 y=4\) \(-3 x+y=5\)

Step-by-Step Solution

Answer

Intersection: (-1, 2). Perpendicular line: \(y = \frac{3}{2}x + \frac{7}{2}\).

Step 1: Solve for Intersections

To find the point of intersection, we need to solve the system of equations given by \(2x + 3y = 4\) and \(-3x + y = 5\). We can use the substitution method to do this. First, solve the second equation for \(y\): \[-3x + y = 5\]\[y = 5 + 3x\]Now substitute \(y = 5 + 3x\) into the first equation:\[2x + 3(5 + 3x) = 4\]\[2x + 15 + 9x = 4\]\[11x = -11\]\[x = -1\]Now substitute \(x = -1\) back into \(y = 5 + 3x\):\[y = 5 + 3(-1)\]\[y = 2\]The point of intersection is \((-1, 2)\).

Step 2: Find the Slope of the First Line

The line equation \(2x + 3y = 4\) can be rewritten in slope-intercept form \(y = mx + c\) to find the slope. Solving for \(y\), we get: \[3y = -2x + 4\]\[y = -\frac{2}{3}x + \frac{4}{3}\]The slope \(m\) of the first line is \(-\frac{2}{3}\).

Step 3: Determine the Perpendicular Slope

The slope of a line perpendicular to another is the negative reciprocal of the other line's slope. Thus, the perpendicular slope of the original line is: \[m_{\perp} = -\left(-\frac{3}{2}\right) = \frac{3}{2}\]

Step 4: Write the Equation for the Perpendicular Line

Now we will use the point-slope form of a line equation \(y - y_1 = m(x - x_1)\) where \((x_1, y_1)\) is the point of intersection \((-1, 2)\). Using the slope \(\frac{3}{2}\), the equation becomes:\[y - 2 = \frac{3}{2}(x + 1)\]Distributing the slope:\[y - 2 = \frac{3}{2}x + \frac{3}{2}\]Solving for \(y\):\[y = \frac{3}{2}x + \frac{3}{2} + 2\]\[y = \frac{3}{2}x + \frac{7}{2}\]Hence, the equation of the line through the intersection point and perpendicular to the original line is \(y = \frac{3}{2}x + \frac{7}{2}\).

Key Concepts

Solving Systems of Equations

The process of solving systems of equations allows us to find where two lines intersect on a graph. In this case, we have two equations representing two lines: \( 2x + 3y = 4 \) and \( -3x + y = 5 \). To find their intersection, we can use the substitution method. This involves solving one equation for a variable and substituting it into the other equation.

• Start by solving the second equation for \( y \): \( y = 5 + 3x \).
• Then substitute \( y = 5 + 3x \) into the first equation: \( 2x + 3(5 + 3x) = 4 \).
• Solve this equation for \( x \). You will find \( x = -1 \).
• Next, substitute \( x = -1 \) back into the equation \( y = 5 + 3x \) to find \( y = 2 \).

This gives the point of intersection \((-1, 2)\). Solving systems of equations can thus reveal crucial points where two conditions or rules hold true simultaneously.

Slope-Intercept Form

Understanding the slope-intercept form is key to graphing lines and identifying their characteristics. Any linear equation can be expressed as \( y = mx + c \), where \( m \) denotes the slope, and \( c \) the y-intercept. Let's see how it applies here.

• Take the equation \( 2x + 3y = 4 \) and rearrange it to find \( y \) in terms of \( x \).
• Subtract \( 2x \) from both sides to get \( 3y = -2x + 4 \).
• Divide every term by 3, giving \( y = -\frac{2}{3}x + \frac{4}{3} \).

Here, \( -\frac{2}{3} \) is the slope \( m \), and \( \frac{4}{3} \) is the y-intercept \( c \). Recognizing these components helps in determining the direction and steepness of the line as well as where it crosses the y-axis.

Point-Slope Form

The point-slope form is useful when you have a point on a line and know its slope. It’s expressed by \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is a known point, and \( m \) is the slope. Here's how we use it to find a perpendicular line in our exercise.

• We already found the intersection at \((-1, 2)\).
• We determined that a line perpendicular to the first line has a slope \( \frac{3}{2} \), which is the negative reciprocal of \( -\frac{2}{3} \).
• Using the point-slope form with the intersection point and the perpendicular slope gives us: \( y - 2 = \frac{3}{2}(x + 1) \).
• Simplify it to \( y = \frac{3}{2}x + \frac{7}{2} \).

Employing the point-slope form in this way allows easy calculation of equations for lines that are perpendicular or parallel to given lines, contributing to our understanding of geometric relationships.