Problem 13 - Elementary Algebra

Question

Find the equation of the line that contains the two given points. Express equations in the form \(A x+B y=C\), where \(A, B\), and \(C\) are integers. (Objective \(1 \mathrm{~b}\) ) \((2,3)\) and \((7,10)\)

Step-by-Step Solution

Answer

The equation of the line is \(7x - 5y = -1\).

Step 1: Find the Slope

The slope \(m\) of the line can be found using the formula \(m = \frac{y_2-y_1}{x_2-x_1}\). Using the points \((2,3)\) and \((7,10)\), substitute \(x_1 = 2\), \(y_1 = 3\), \(x_2 = 7\), and \(y_2 = 10\) into the formula: \[ m = \frac{10-3}{7-2} = \frac{7}{5} \].

Step 2: Use Point-Slope Form

With the slope \(m = \frac{7}{5}\) and point \((2,3)\), use the point-slope form of a line equation: \(y - y_1 = m(x - x_1)\). Substituting the values gives: \[ y - 3 = \frac{7}{5}(x - 2) \].

Step 3: Convert to Slope-Intercept Form

Simplify the equation from Step 2: \[ y - 3 = \frac{7}{5}x - \frac{14}{5} \]. Adding 3 to both sides to isolate \(y\): \[ y = \frac{7}{5}x - \frac{14}{5} + \frac{15}{5} \], resulting in \[ y = \frac{7}{5}x + \frac{1}{5} \].

Step 4: Convert to Standard Form

To convert to the standard form \(Ax + By = C\), first eliminate the fraction by multiplying through by 5: \[ 5y = 7x + 1 \]. Then rearrange the terms: \[ 7x - 5y = -1 \].

Step 5: Adjust Integers for Standard Form

Ensure the coefficients are integers and the leading coefficient is positive. The equation \(7x - 5y = -1\) fits this requirement. Thus, the standard form of the line equation is \(7x - 5y = -1\).

Key Concepts

Slope

The **slope** of a line is a measure of its steepness and direction. It's one of the first things you need to calculate when finding the equation of a line. The formula to calculate the slope is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]With this formula, you're essentially finding the change in the vertical values (rise) divided by the change in the horizontal values (run) between two points. For example, using the points

• \((2,3)\) and \((7,10)\)

you substitute into the formula:\[ m = \frac{10-3}{7-2} = \frac{7}{5} \] This result, \(\frac{7}{5}\), indicates that for every 5 units you move horizontally, the line moves 7 units vertically.

Point-Slope Form

**Point-slope form** is a quick way to get to an equation using the slope and any point on the line other than the y-intercept. This form is expressed as:\[ y - y_1 = m(x - x_1) \]Where \( m \) is the slope, and \( (x_1, y_1) \) is a known point on the line. Let's use our slope \( \frac{7}{5} \) and a point, \( (2, 3) \):

• Substitute into the formula: \[ y - 3 = \frac{7}{5}(x - 2) \]

This equation now represents the line. If you have a different point, you could use this same form.Point-slope form is particularly useful when you're given two points and need to quickly write the equation of a line.

Standard Form

In **standard form**, a line equation is expressed as \( Ax + By = C \), where \( A, B, \) and \( C \) are integers. This form is useful for solving systems of equations and analyzing linear relationships. Converting from slope-intercept to standard form begins with eliminating fractions:Given from previous steps, \( y = \frac{7}{5}x + \frac{1}{5} \).

• Multiply through by 5 to clear fractions: \[ 5y = 7x + 1 \]

Then rearrange to get:

• \[ 7x - 5y = -1 \]

Make sure \( A \), the coefficient of \( x \), is positive, and all coefficients are integers. In this case, \( 7x - 5y = -1 \) satisfies these criteria. This neatly structured form makes handling equations in algebraic applications straightforward.