Linear Functions

6.5 Slope-Point Form of the Equation for a Linear Function

Today’s Objectives

• Relate linear relations expressed in: slope-intercept form, general form, and slope-point form to their graphs, including:• Express a linear relation in different forms, and

compare graphs• Rewrite a linear relation in either slope-intercept or

general form• Graph, with or without technology, a linear relation in

slope-intercept, general, or slope-point form• Identify equivalent linear relations from a set of linear

relations• Match a set of linear relations to their graphs

Equations of a Linear Function

• We can make an equation that describes a line’s location on a graph. This is called a linear equation. There are three forms of linear equation that we will be looking at:

• Standard Form: Ax + By + C = 0, where A, B, and C are integers.

• Slope y-intercept form: y = mx + b, where m is the slope, and b is the y-intercept.

• Slope-point form: y – y1 = m(x – x1), where m is the slope, and the line passes through a point located at (x1, y1)

Slope-point form

• When we know the slope of a line and the coordinates of a point on the line, we use the property that the slope of a line is constant to determine an equation for the line.

• In this example, the line has a slope of -3 and passes through . We use any other point on the line to write an equation for the slope, m:

(-2,5)

ExampleA line has a slope of -3 and passes through P(-2,5). Write an equation in slope-point form.

Slope = rise/runm = y - 5 / x - (-2)-3 = y - 5 / x + 2-3(x+2) = y – 5So y – 5 = -3(x+2) {slope-point form}

Example: • Describe the graph of the equation in slope-point form,

then graph the equation

• Solution:

From the equation, we can see the x-value of the point P, is at -4, and the y coordinate of the point P is at 2. So, point P is found at (-4, 2). Also, we can see that the slope is 1/3. We can use this information to now graph the line.

Example

Writing an Equation Using a Point on the Line and its Slope• A) Write an equation in slope-point form for this line:• B) Write the equation in part A in slope-intercept form. What

is the y-intercept of this line?

Example 2 - Solution• A) Count out the rise and run to determine the slope of the

line.

• Use the slope-point form of the equation:• Substitute a point easily read from the graph and our slope: (-

1,-2):

• B) First, we can remove the brackets by multiplying:

•

• Next, we solve for y:• As we can see from this equation, the y-intercept is at:

Parallel and Perpendicular Lines• Writing the equations of parallel and perpendicular lines in

slope-point form is very similar to doing the same in slope-intercept form.

• The only difference between the equations is in the value of m, the slope.

• Recall that the slope of a line perpendicular to another line is the negative reciprocal of the original line.

Example

Write an equation for the line that passes through R(1,-1) and is:• A) Parallel to the line • B) Perpendicular to the line

Solution:• First, we can see from the lines that the slope is 2/3. Any line

parallel to this line has slope 2/3. Now, we use our point R(1,-1) to find the equation of the line:

• Any line perpendicular to this line has a slope that is the negative reciprocal of 2/3, which is -3/2. So, the equation of a line perpendicular to the line is:

Example:Write an equation for the line that passes through S(2,-3) and is:• A) Parallel to the line y = 3x + 5• B) Perpendicular to the line y = 2/3x - 5

Solution:• First, we can see from the lines that the slope is 3. Any line parallel

to this line has slope 3. • Now, we use our point to find the equation of the line: y – (-3) = 3(x – 2)

y + 3 = 3(x-2)

• Any line perpendicular to this line has a slope that is the negative reciprocal of 3, which is -1/3. So, the equation of a line perpendicular to the line is:

y + 3 = -1/3(x-2)

Using 2 Points to Write an Equation

• We can use the coordinates of two points that satisfy a linear function P(x1,y1) and Q(x2,y2) to write an equation for the function.

• We write the slope of the graph of the function in two ways:

• So, an equation is:

ExampleGiven K(1,-2) and J(6,8)Determine the slope-point form of the equation.

Calculate the slope of the line.m = 8-(-2)/6-1 m = 2

Use either point K or J to write the slope-intercept form of the equation.Using K(1,-2) Using J(6,8)y – (-2) = 2(x -1) y + 2 = 2(x -1) y – 8 = 2(x -6)

y = 2x - 2 – 2 y = 2x – 12 + 8y = 2x – 4 y = 2x – 4

Example

Write the equation of a Linear Function Given Two Points• The sum of angles, s degrees, in a polygon is a linear function

of the number of sides, n, of the polygon. The sum of the angles in a triangle (3 sides) is 180º. The sum of the angles in a quadrilateral (4 sides) is 360º.

• A) Write the linear equation to represent this function• B) Use the equation to determine the sum of angles in a

dodecagon (shape with 12 sides)

HINT: Determine your x and y axis ? ? ? ? ? number of side [n] is the independent variable (x) sum of angles [s] is the dependent variable (y)

Determine the coordinates of the two points given.Determine the slope.

Example 4

Solution:• A) Slope-point form:

• Slope y-intercept form:

• B)

• The sum of the angles in a dodecagon is 1800º