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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. \(2 x+3 y=4\) \(-3 x+y=5\)

Short Answer

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

Step by step solution

01

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 \((-1, 2)\).
02

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 \(-\frac{2}{3}\).
03

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: \[m_{\perp} = -\left(-\frac{3}{2}\right) = \frac{3}{2}\]
04

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 \(y = \frac{3}{2}x + \frac{7}{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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