Problem 90 - Algebra for College Students

Question

Find the equation of line l in each case and then write it in standard form with integral coefficients. Line \(l\) goes through \((-1,-2)\) and is perpendicular to \(y=-3 x+7\).

Step-by-Step Solution

Answer

The equation of the line is: \[ x - 3y = 5 \]

Step 1 - Identify the slope of the given line

The given line's equation is in the slope-intercept form: \[ y = -3x + 7 \]From this, we can see the slope of this line (m) is -3.

Step 2 - Find the slope of the perpendicular line

If two lines are perpendicular, their slopes are negative reciprocals of each other. Therefore, the slope of the perpendicular line will be \[ m_{\text{perpendicular}} = -\frac{1}{-3} = \frac{1}{3} \]

Step 3 - Write the equation in point-slope form

Using the point-slope form equation \[ y - y_1 = m (x - x_1) \]where \( m = \frac{1}{3} \) and the point \( (-1, -2) \), we have \[ y + 2 = \frac{1}{3} (x + 1) \]

Step 4 - Simplify to slope-intercept form

Expanding and simplifying the equation, we get: \[ y + 2 = \frac{1}{3}x + \frac{1}{3} \]Subtract 2 from both sides: \[ y = \frac{1}{3}x + \frac{1}{3} - 2 \]Combine terms: \[ y = \frac{1}{3}x - \frac{5}{3} \]

Step 5 - Convert to standard form

To convert to standard form with integral coefficients, multiply every term by 3 to clear the fractions: \[ 3y = x - 5 \]Rearrange to the standard form, \[ Ax + By = C \]So, we get: \[ -x + 3y = -5 \]For positive coefficients, multiply by -1: \[ x - 3y = 5 \]

Key Concepts

slope-intercept form

The slope-intercept form of a line's equation is one of the most common and useful forms. It is written as \[ y = mx + b \] where:

• \( m \) is the slope of the line, which indicates the steepness and direction.
• \( b \) is the y-intercept, which is where the line crosses the y-axis.

For example, the line given in the exercise has an equation \[ y = -3x + 7 \] Here, the slope \( m \) is -3, and the y-intercept \( b \) is 7. This tells us that for every 1 unit we move to the right along the x-axis, the line will move down 3 units. The y-intercept is the point (0,7). This form is particularly handy for quickly sketching lines and understanding their behavior.

point-slope form

The point-slope form is another useful way to write the equation of a line. It is written as \[ y - y_1 = m (x - x_1) \] where:

• \( (x_1, y_1) \) is a specific point on the line.
• \( m \) is the slope of the line.

This form is particularly useful when you have one point on the line and the slope. For example, from the exercise, we used the point (-1,-2) and the slope \( \frac{1}{3} \), which is the negative reciprocal of -3. Plugging in these values, we get: \[ y + 2 = \frac{1}{3} (x + 1) \] This tells us that the line goes through the point (-1,-2) and has a slope of \( \frac{1}{3} \).

standard form

The standard form of a line's equation is written as \[ Ax + By = C \] where:

• \( A \), \( B \), and \( C \) are integers.
• \( A \) and \( B \) are not both zero.

The standard form is useful for certain types of algebraic operations, like systems of equations. To convert a line equation to standard form, you need to rearrange terms and clear any fractions. From the exercise solution, the slope-intercept form \( y = \frac{1}{3}x - \frac{5}{3} \) was converted to standard form by:

• Multiplying every term by 3 to clear fractions: \( 3y = x - 5 \).
• Rearranging to \( -x + 3y = -5 \).
• Converting to positive coefficients: \( x - 3y = 5 \).

This final form makes it easier to understand and to use in further calculations.