Problem 21 - Elementary Algebra

Question

For Problems \(13-22\), 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. \((0,4)\) and \((7,0)\)

Step-by-Step Solution

Answer

The equation of the line is \(4x + 7y = 28\).

Step 1: Determine the slope of the line

To find the slope, use the formula \( m = \frac{y_2-y_1}{x_2-x_1} \). Here, \( (x_1, y_1) = (0,4) \) and \( (x_2, y_2) = (7,0) \). Substitute the values: \[ m = \frac{0 - 4}{7 - 0} = \frac{-4}{7} \]. So, the slope \( m \) is \( -\frac{4}{7} \).

Step 2: Use the point-slope form of the equation

The point-slope form of an equation is \( y - y_1 = m(x - x_1) \). We will use the point \((0, 4)\) and the slope \(-\frac{4}{7}\). Substitute into the formula: \[ y - 4 = -\frac{4}{7}(x - 0) \]. Simplify this to get: \[ y - 4 = -\frac{4}{7}x \].

Step 3: Convert to standard form

Convert the equation \( y = -\frac{4}{7}x + 4 \) into standard form \( Ax + By = C \). Multiply through by 7 to eliminate the fraction: \[ 7y = -4x + 28 \]. Rearrange terms to get: \[ 4x + 7y = 28 \].

Step 4: Verify integer coefficients

In the equation \(4x + 7y = 28\), the coefficients \( A = 4 \), \( B = 7 \), and \( C = 28 \) are already integers. Therefore, the equation is in the correct form.

Key Concepts

Slope Calculation

The slope of a line describes how steep the line is and in which direction it tilts. To calculate the slope between two points, we use the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]where

• \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points on the line.
• \(m\) represents the slope.

In this exercise, we need to find the slope between the points \((0,4)\) and \((7,0)\). Applying these coordinates to the formula gives us:\[ m = \frac{0 - 4}{7 - 0} = \frac{-4}{7} \]This result \(m = -\frac{4}{7}\) means our line slopes downwards from left to right, as it is negative. This understanding of slope is crucial, because the formula reveals how much \(y\) changes for each unit of change in \(x\). A negative slope like \(-\frac{4}{7}\) shows that for every 7 units you move horizontally (along the \(x\)-axis), \(y\) decreases by 4 units.

Standard Form of Linear Equation

The standard form of a linear equation is represented as:\[ Ax + By = C \] where

• \(A\), \(B\), and \(C\) are integers,
• the equation contains no fractions, and
• \(A\) should ideally be positive.

In our solution, after determining the slope, we rewrote the line equation in this standard form. To do so, we started from a simplified point-slope form \(y = -\frac{4}{7}x + 4\) and cleared the fractional coefficient by multiplying each term by 7:\[ 7y = -4x + 28 \]By rearranging, we ensure \(x\) and \(y\) terms are on one side:\[ 4x + 7y = 28 \]Notice how this conversion leads us to an equation where \(4\) and \(7\) are the coefficients of \(x\) and \(y\) respectively, and the equation equals \(28\). These are all integers, fitting our need for the standard form.

Point-Slope Form

The point-slope form is a useful way to write the equation of a line when you know a point on the line and the line’s slope. It is expressed as:\[ y - y_1 = m(x - x_1) \]Here:

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

In our exercise, we applied the point-slope form using the point \((0, 4)\) and the earlier calculated slope \(-\frac{4}{7}\). Plugging these values into the formula gives:\[ y - 4 = -\frac{4}{7}(x - 0) \]Simplifying gives:\[ y - 4 = -\frac{4}{7}x \]This format makes it convenient to see how each change in \(x\) affects \(y\). Once you have understood the point-slope form, you can transition smoothly to other forms of the linear equation, like the slope-intercept or standard form, depending on what the problem requires.