Problem 74 - Algebra for College Students

Question

The equation of a line that contains the two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) is \(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\). We often refer to this as the two-point form of the equation of a straight line. Use the two- point form and write the equation of the line that contains each of the indicated pairs of points. Express final equations in standard form. (a) \((1,1)\) and \((5,2)\) (b) \((2,4)\) and \((-2,-1)\) (c) \((-3,5)\) and \((3,1)\) (d) \((-5,1)\) and \((2,-7)\)

Step-by-Step Solution

Answer

(a) \(x - 4y = -3\); (b) \(5x - 4y = -6\); (c) \(2x + 3y = 9\); (d) \(8x + 7y = -33\).

Step 1: Use the Two-Point Form Formula

Start with the given formula for the two-point form: \[ \frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \] For each pair of points, substitute the coordinates of the points into this formula.

Step 2: Substitute for Pair (a) (1,1) and (5,2)

With points (1,1) and (5,2), the formula becomes:\[ \frac{y-1}{x-1} = \frac{2-1}{5-1} = \frac{1}{4} \]Multiply through by \(x-1\) to clear the fraction:\[ y-1 = \frac{1}{4} (x-1) \].Simplify and solve for \(y\) in terms of \(x\):\[ y = \frac{1}{4}x + \frac{3}{4} \].

Step 3: Convert Equation (a) to Standard Form

The equation from Step 2 is \[ y = \frac{1}{4}x + \frac{3}{4} \]. To get it into standard form, multiply everything by 4 to eliminate the fraction:\[ 4y = x + 3 \].Rearrange this to get:\[ x - 4y = -3 \]. This is the standard form \(Ax + By = C\).

Step 4: Substitute for Pair (b) (2,4) and (-2,-1)

For points (2,4) and (-2,-1), use the formula:\[ \frac{y-4}{x-2} = \frac{-1-4}{-2-2} = \frac{-5}{-4} = \frac{5}{4} \].Clear the fraction by multiplying by \(x-2\):\[ y-4 = \frac{5}{4} (x-2) \].Simplify to:\[ y = \frac{5}{4}x - \frac{5}{2} + 4 \].Combine terms:\[ y = \frac{5}{4}x + \frac{3}{2} \].

Step 5: Convert Equation (b) to Standard Form

The equation is \( y = \frac{5}{4}x + \frac{3}{2} \). Multiply through by 4 to clear the fractions:\[ 4y = 5x + 6 \].Rearrange to:\[ 5x - 4y = -6 \].

Step 6: Substitute for Pair (c) (-3,5) and (3,1)

For points (-3,5) and (3,1):\[ \frac{y-5}{x+3} = \frac{1-5}{3+3} = \frac{-4}{6} = \frac{-2}{3} \].Clear the fractions by multiplying:\[ y-5 = -\frac{2}{3}(x+3) \].Simplify to:\[ y = -\frac{2}{3}x - 2 + 5 \].Combine terms:\[ y = -\frac{2}{3}x + 3 \].

Step 7: Convert Equation (c) to Standard Form

The equation is \( y = -\frac{2}{3}x + 3 \). Multiply by 3 to clear fractions:\[ 3y = -2x + 9 \].Rearrange to:\[ 2x + 3y = 9 \].

Step 8: Substitute for Pair (d) (-5,1) and (2,-7)

For points (-5,1) and (2,-7):\[ \frac{y-1}{x+5} = \frac{-7-1}{2+5} = \frac{-8}{7} \].Clear fractions by multiplying:\[ y-1 = -\frac{8}{7}(x+5) \].Simplify to:\[ y = -\frac{8}{7}x - \frac{40}{7} + 1 \].Combine terms:\[ y = -\frac{8}{7}x - \frac{33}{7} \].

Step 9: Convert Equation (d) to Standard Form

The equation is \( y = -\frac{8}{7}x - \frac{33}{7} \). Multiply everything by 7:\[ 7y = -8x - 33 \].Rearrange to:\[ 8x + 7y = -33 \].

Key Concepts

Standard Form Equations

Standard form equations are a way to express linear equations. The typical structure is \( Ax + By = C \), where \( A \), \( B \), and \( C \) are integers. To express any linear equation in this form, you simply need to rearrange the terms from a slope-intercept or another linear form.
This form is especially useful because:

• The coefficients \( A \), \( B \), and \( C \) give clear information at a glance.
• It assists in finding intercepts quickly; for instance, setting \( x = 0 \) finds the \( y \)-intercept as \( C/B \), and \( y = 0 \) finds the \( x \)-intercept as \( C/A \).
• It is often used in conjunction with methods for solving systems of equations.

Converting an equation like \( y = \frac{1}{4}x + \frac{3}{4} \) involves eliminating fractions and rearranging terms, resulting in forms like \( x - 4y = -3 \).
Remember, the standard form is a fundamental tool in algebra, particularly when dealing with coordinate geometry problems.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, is the study of geometry using a coordinate system. This branch of mathematics allows us to use algebraic methods to solve geometric problems using coordinates on a graph.
Key components in coordinate geometry include:

• Points: Defined by ordered pairs \( (x, y) \) which indicate a location on the plane.
• Lines: Defined by equations like the two-point form, which lets us connect any two points.
• Slopes: This measures the steepness of a line, calculated as \( \frac{y_2-y_1}{x_2-x_1} \).

When presented with two points, coordinate geometry enables us to find the exact line equation by substituting the point values into forms like the two-point form. This mathematical framework provides precise tools for describing geometric shapes and their positions in the plane.

Linear Equations

Linear equations are equations of the first degree, meaning they involve variables raised to the power of one. In their simplest form, they are written as \( y = mx + c \), where \( m \) represents the slope, and \( c \) is the y-intercept. However, they can be expressed in various forms, including the standard form \( Ax + By = C \).
Key characteristics of linear equations include:

• A straight line graph.
• A constant rate of change, represented by a slope \( m \).
• Solutions that create linear representations on a graph, making it easy to understand and predict.

When solving exercises involving linear equations, such as finding the equation through two points, the process involves calculating the slope between points using \( m = \frac{y_2-y_1}{x_2-x_1} \) and then substituting these into the equation form, either slope-intercept or two-point.
This versatility in forms allows for varied approaches to solve real-world problems using linear equations.