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2. Equations & Inequalities in Two Variables
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Integrated Math 1

2.03 Point-slope form

Lesson
Practice
Lesson

Concept summary

When we are given the coordinates of a point on the line and the slope of that line, then point-slope form can be used to state the equation of the line.

Coordinates can be given as an ordered pair, in a table of values, read from a graph, or described in a scenario. The slope can be stated as a value, calculated from two points, read from a graph, or given as a rate of change in a scenario.

Point-slope form of a linear relationship:

\displaystyle y-y_1=m\left(x-x_1\right)
\bm{m}
The slope of the line
\bm{x_1}
The x-coordinate of the given point
\bm{y_1}
The y-coordinate of the given point
Coordinate

A number used to locate a point on a number line. One of the numbers in an ordered pair, or triple, that locates a point on a coordinate plane or in coordinate space, respectively.

Ordered pair

A point on a graph written as \left(x, y\right). Also called coordinates, or a coordinate pair.

We can also find the equation in point-slope form when given the coordinates of two points on the line, by first finding the slope of the line.

We now have three forms of expressing the same equation, and each provides us with useful information about the line formed.

Point-slope form is useful when we know, or want to know:

  • The slope of the line
  • A point on the line

Slope-intercept form is useful when we know, or want to know:

  • The slope of the line
  • The y-intercept of the line

Standard form is useful when we know, or want to know:

  • Both intercepts of the line

Worked examples

Example 1

For each of the following equations, determine if they are in point-slope form, standard form, or slope-intercept form. If they are not in standard form, convert them to standard form.

a

y-3=-\dfrac{2}{5}\left(x+7\right)

Solution

This is in point-slope form.

\displaystyle y-3\displaystyle =\displaystyle -\dfrac{2}{5}\left(x+7\right)State the given equation
\displaystyle 5y-15\displaystyle =\displaystyle -2\left(x+7\right)Multiplication property of equality
\displaystyle 5y-15\displaystyle =\displaystyle -2x-14Distributive property
\displaystyle 2x+5y\displaystyle =\displaystyle 1Addition property of equality

2x+5y=1 is in standard form.

Reflection

There are more steps to convert from point-slope form to standard form, than from point-slope form to slope-intercept form.

b

y=4x-10

Solution

This is in slope-intercept form.

\displaystyle y\displaystyle =\displaystyle 4x-10State the given equation
\displaystyle -4x+y\displaystyle =\displaystyle -10Subtraction property of equality
\displaystyle 4x-y\displaystyle =\displaystyle 10Division property of equality

4x-y=10 is in standard form.

Example 2

A line passes through the two points \left(-3,7\right) and \left(2,-3\right). Write the equation of the line in point-slope form.

Approach

We will first find the slope of the line using the two points. Then we will pick one of the points to substitute into the point-slope equation.

Solution

Find the slope of the line:

\displaystyle m\displaystyle =\displaystyle \dfrac{y_2-y_1}{x_2-x_1}Formula for slope of a line
\displaystyle =\displaystyle \dfrac{-3-7}{2-\left(-3\right)}Substitute in the points
\displaystyle =\displaystyle \dfrac{-10}{5}Simplify the subtraction
\displaystyle =\displaystyle -2Simplify the fraction

Find the equation of the line in point-slope form:

\displaystyle y-y_1\displaystyle =\displaystyle m\left(x-x_1\right)Formula for point-slope form
\displaystyle y-7\displaystyle =\displaystyle -2\left(x-\left(-3\right)\right)Substitute known values into the formula
\displaystyle y-7\displaystyle =\displaystyle -2\left(x+3\right)Simplify

Reflection

We can confirm we have correctly written this in point-slope form: y-y_1=m\left(x-x_1\right), as we can see the coordinates \left(-3,7\right), and a slope of -2 are represented correctly. This is easier to see in the previous line: y-7=-2\left(x-\left(-3\right)\right)

Example 3

A carpenter charges for a day's work using the given equation, where y is the cost and x is the number of hours worked:y-125=50\left(x-2\right)

a

Draw the graph of the linear equation from point-slope form. Clearly label the axes with labels, units, and an appropriate scale.

Approach

We need two points to plot a line. We can read from the equation that one point on the line is \left(2,125\right). To get another point, we can use the slope from the given point or substitute in an x-value in the domain and solve for y.

Since the x-axis will be the number of hours worked and generally an 8 hour work day is reasonable, showing the graph for 0\leq x\leq 8 would be a good scale.

Solution

Before we graph, we label our axes, to do that we need to know what our maximum and minimum values should be.

We can use the slope or the equation to determine the y-value when x=8. We can find that the y-value when x=8 is y=425.

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\text{Hours }(x)
50
100
150
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450
\text{Charge in }\$ \,(y)

A reasonable scale for the x-axis from 0 to 8 going up by 1.

A reasonable scale for the y-axis from 0 to 450 going up by 50 with one tick between each label at the 25s.

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\text{Hours } (x)
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\text{Charge in }\$ \,(y)

The given point is \left(2,125\right).

Since the slope is 50, we can go up 50 units and right 1 unit from our given point to plot another point on the line.

b

Predict the charge for 6 hours of work using the graph.

Approach

We can go to x=6 on the x-axis and then go up to the line and across to the y-axis to make our prediction.

Solution

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\text{Hours } (x)
50
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\text{Charge in }\$ \,(y)

The charge for 6 hours of work will be \$ 325.

c

Give an example of a non-viable solution if the carpenter only uses this model for a maximum of 10 hours per day. Explain your answer.

Approach

Since the carpenter only uses this model between 0 and 10 hours, any solution outside of this interval will not be viable.

Solution

One possible non-viable solution would be \left(11, 575\right). Which would represent working 11 hours and getting paid \$575, but the model only applies to a maximum of 10 hours, so we don't actually know what would happen for 11 hours.

Reflection

Any solution with x<0 or x>10 would be non-viable.

Outcomes

M1.N.Q.A.1

Use units as a way to understand real-world problems.*

M1.N.Q.A.1.A

Choose and interpret the scale and the origin in graphs and data displays.*

M1.A.SSE.A.2

Use the structure of an algebraic expression to identify ways to rewrite it.

M1.A.CED.A.2

Create equations in two variables to represent relationships between quantities and use them to solve problems in a real-world context. Graph equations with two variables on coordinate axes with labels and scales, and use the graphs to make predictions.*

M1.A.CED.A.3

Create individual and systems of equations and/or inequalities to represent constraints in a contextual situation, and interpret solutions as viable or non-viable.*

M1.MP1

Make sense of problems and persevere in solving them.

M1.MP2

Reason abstractly and quantitatively.

M1.MP3

Construct viable arguments and critique the reasoning of others.

M1.MP4

Model with mathematics.

M1.MP5

Use appropriate tools strategically.

M1.MP6

Attend to precision.

M1.MP7

Look for and make use of structure.

M1.MP8

Look for and express regularity in repeated reasoning.

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